hardware in loop interview questions

Hiring a hardware engineer can be a daunting task, especially if you are unfamiliar with the hardware engineering field. To ensure you make the best hire for your organization, it’s important to ask the right questions during your interviews. Naturally, the types of questions you should ask will vary depending on the type of position you are looking to fill. If you’re looking to hire a hardware engineer with experience in hardware-in-loop (HIL) systems, it’s important that you understand the related interview questions. This blog post will provide a comprehensive overview of hardware-in-loop interview questions, including the types of questions to ask and tips for structuring a successful interview. We’ll also explore the key topics related to hardware-in-loop engineering, so you can be sure you’re asking the right questions to assess each applicant’s expertise. So, whether you’re seeking a hardware-in-loop engineer for your

The Interview: Hardware in the Loop (HIL)
  • Can you introduce yourself in a few words? …
  • Can you explain the acronym HIL? …
  • Can you also quickly explain MIL, SIL, DIL ? …
  • Is there an alternative to HiL test benches? …
  • What types of problems can HIL test benches solve?

Hardware in the Loop Testing

Interview Questions of Top Designations

hardware in loop interview questions

hardware in loop interview questions

hardware in loop interview questions

Difference in their Usage:

While a union allows us to treat the same space in memory as a number of different variables, a structure only allows us to treat a number of different variables stored at different locations in memory. In other words, a union provides a means of treating a portion of memory as one type of variable on occasion and as a different type of variable on another occasion. When working with hardware, it is frequently necessary to access a byte or group of bytes at once, and occasionally each byte separately. Usually union is the answer. =======Difference With example=======.

Consider creating a structure with an int, char, and float, then declaring a union with those three types of data.

Size of TT (struct) would be >9 bytes (compiler dependent-if int,float, and char are taken as 4,4,1) Size of UU (Union) would be 4 bytes as supposed from above If a double variable is present in the union, the size of the union and the struct would be 8 bytes, as would the total size of all the variables in the struct.

I have a two-day virtual on-site interview scheduled for Meta. Two coding rounds and one behavioral round are already complete for me. I unnecessarily complicated the code in the first round of coding, which prevented me from finishing the solution. I gave a verbal explanation and fixed some bugs. The second one was decent; I accidentally left some unnecessary code in during code cleanup, but I was still able to provide the best solution and respond to follow-up inquiries about how I could use less memory. I have design rounds today, and after the first coding round, I feel very unmotivated for these rounds. It was a moderately simple question, but I believe that because of my stress while coding, I didn’t fully consider it. Just curious about my chances here and whether or not doing well in design will help me turn things around. Current TC – 180k Location – Bay Area.

Detailed Example:

struct foo { char c; long l; char *p; };

union bar { char c; long l; char *p; };

All of the components of a struct foo are c, l, and p. Each element is separate and distinct.

One of the elements c, l, or p is always present in a union bar. Since each element starts at the same memory location and is stored there, you can only refer to the element that was most recently stored. (For example, after “barptr->c = 2,” referencing any other elements, such as “barptr->p,” would result in behavior that is not defined. ).

Try the following program. (Yes, I am aware that this causes the aforementioned “undefined behavior,” but most computers will still produce some sort of output. ).

struct foo { char c; long l; char *p; };

union bar { char c; long l; char *p; };

union bar mybar; struct foo myfoo; int main(int argc,char *argv[])

myfoo. c = 1; myfoo. l = 2L; myfoo. p = “This is myfoo”;.

mybar. c = 1; mybar. l = 2L; mybar. p = “This is mybar”;.

printf(“myfoo: %d %ld %s ”,myfoo.c,myfoo.l,myfoo.p); printf(“mybar: %d %ld %s ”,mybar.c,mybar.l,mybar.p);

On my system, I get:

Mybar: 100 4197476 This is Mybar myfoo: 1 2 This is Myfoo ===========================================

Predefined data types or other types of structure can be included in a structure. The total length of the structure’s components determines its length or size. In C, structures cannot contain functions. in C++ it can.

Union: A union is a combination of elements. These elements can be other unions or predefined data types. However, the union’s size and length are its maximum internal components.

Because of padding, which again depends on OS, the sizeof() operator returns a size that is slightly larger than the calculated size. Unions allocate memory according to the maximum amount needed by each member, whereas structures allocate memory according to the total amount needed by all members. In a union, each member uses a single block, whereas in a structure, each member has a separate memory space.

21. What is meant by structure padding?

Answer: compilers pad structures to optimize data transfers. This is an hardware architecture issue. The majority of modern CPUs operate most efficiently when basic types, like “int” or “float,” are aligned on memory boundaries of a specific size (e.g. often a 4byte word on 32bit archs). Many architectures forbid misaligned access or impose a performance penalty if it does occur. In order to meet alignment requirements, a compiler will add extra bytes between fields when processing a structure declaration.

Most processors require specific memory alignment on variables certain types. The size of the relevant basic type is typically the minimum alignment; for example, this is common.

Char variables can appear at any byte boundary and can be byte aligned.

Short (2 byte) variables can appear at any even byte boundary, but they must be 2 byte aligned. It follows that while 0x10004566 is a valid location for a short variable, 0x10004567 is not.

In order to appear at byte boundaries that are a multiple of 4 bytes, long (4 byte) variables must be 4 byte aligned. Accordingly, 0x10004566 is an invalid location for a long variable, whereas 0x10004568 is.

Structure padding is necessary because the structure’s members must appear at the right byte boundary; to do this, the compiler inserts padding bytes (or bits, if bit fields are being used) before the structure’s members are displayed. Additionally, the structure’s size must ensure that every structure in an array of structures is correctly aligned in memory, which allows for the possibility of padding bytes at the structure’s end.

Example of a struct: char c1, short s1, char c2, long l1, char c3,

If the alignment pattern I previously stated holds true in this structure, then

There must be a padding byte between c1 and s1 because c1 can appear at any byte boundary while s1 must appear at a 2 byte boundary.

When l1 is at a 4 byte boundary, c2 can then appear in the available memory location, but there are 3 padding bytes between c2 and l1.

c3 will then appear in the free space in memory, but due to the structure’s inclusion of a long member, the structure’s size and alignment must both be multiples of 4 bytes. Therefore, the structure ends with 3 padding bytes. It would appear in memory in this order.

S1 Byte 1 S1 Byte 2 C2 Byte Padding Padding Padding L1 Byte 1 L1 Byte 2 L1 Byte 3 L1 Byte 4 C3 Padding Padding Padding Padding Padding Byte

The structure would be 16 bytes long.

Example of a structure: long l1, short s1, char c1, char c2, char c3,

There will be no need for padding between l1 and s1 because l1 will then appear with the proper byte alignment and s1 will be correctly aligned. c1, c2, c3 can appear at any location. Since the structure must be a multiple of 4 bytes in size because it contains a long, 3 padding bytes follow c3 in memory.

s1 byte 1 s1 byte 2 c1 c2 c3 padding byte padding byte padding byte l1 byte 1 l1 byte 2 l1 byte 3 l1 byte 4

and is only 12 bytes long.

I should point out that structure packing depends on the platform, compiler, and sometimes the compiler switch.

Memory pools are merely a portion of memory set aside for temporary allocation to other application components.

A memory leak happens when memory is allocated from the heap (or a pool) and then all references to that memory are deleted without the memory being returned to the pool from which it was allocated.


  • struct MyStructA {
  • char a;
  • char b;
  • int c;
  • };
  • struct MyStructB {
  • char a;
  • int c;
  • char b;
  • };
  • int main(void) {
  • int sizeA = sizeof(struct MyStructA);
  • int sizeB = sizeof(struct MyStructB);
  • printf(“A = %d ”, sizeA);
  • printf(“B = %d ”, sizeB);
  • return 0;
  • }
  • 22. What distinguishes macro variables from constant variables in C?

    Preprocessors replace macros, but compilers check for constant data types when using them When a programmer wants to change values only in a single function, they occasionally replace macros without checking the values and instead prefer to use constants.

    The first technique comes from the C programming language. The preprocessor directive #define can be used to define constants. The preprocessor is a program that modifies your source file before compilation. Preprocessor directives like #include, #define, and #if/#endif are frequently used to conditionally control which parts of your code will be compiled. #include is used to include additional code into your source file. The #define directive is used as follows.

    #define pi 3.1415#define id_no 12345

    In your source file, the preprocessor replaces the constant with its value wherever it appears. Thus, every “pi” in your source code will be changed to 3 for example. 1415. The compiler will only see the value 3. 1415 in your code, not “pi”. This technique has a flaw in that the replacement is performed lexically without type checking, bound checking, or scope checking. Every “pi” is just replaced by its value. The method needs to be avoided because it is out-of-date and only supports legacy code. Const The second method is to define a variable with the keyword const. Any attempts to modify variables that have been declared const will be detected by the compiler when in use.

    const float pi = 3.1415;const int id_no = 12345;

    There are two main advantages over the first technique.

    First, the type of the constant is defined. “pi” is float. “id_no” is int. This allows some type checking by the compiler. Second, these constants are variables with clear boundaries. A variable’s scope relates to the areas of your program where it is defined. Some variables might only be present in specific functions or code blocks.

    You might want to use “id_no” in one function and a totally unrelated “id_no” in your main program, for instance.

    23. What distinguishes a C recursive function from a re-entrant function?

    Re entrant functions are functions that are guaranteed to operate effectively in multi-threaded environments. while a function is accessed by one thread, another thread may call it, indicating that each thread has its own execution stack and handling. Therefore, no static or shared variables that could harm or interfere with the execution should be present in the function. Means a function that can run safely and correctly from one thread while being called by another.


    Recursive function Example:

    doll (size – 1); // Decrements the size variable so the next doll will be smaller doll (size == 0); // No doll can be smaller than 1 atom (100==1) so doesn’t call itself return; // Return does not have to return anything; it can be used to exit a function. The program begins with a big doll (10); it uses a logarithmic scale.

    24. What is V-Model ? What are the benefits?

    The V-model is a software development process that is similar to the waterfall model but also applicable to hardware development. After the coding stage, the process steps are bent upwards to create the familiar V shape rather than moving down the line. The V-Model depicts the connections between each stage of the development life cycle and the corresponding testing stage.

    The V model has a number of benefits:

    1. A test approach, or test strategy document, which specifies how testing will be carried out throughout the project’s lifecycle, is typically present in systems development projects. Part of that strategy has a consistent foundation and benchmark thanks to the V model.

    2. The V model specifically advises that testing (quality assurance) be taken into account early on in a project’s life. At any point during the lifecycle, testing and fixing can be done. However, as development continues, the price of locating and rectifying errors rises sharply. Evidence demonstrates that if a design flaw costs $1. 0 dollars to fix, but the same error discovered just before testing will cost 6 dollars. 5 units, 15 units during testing, and 60 to 100 units after release The requirement for quality assurance of documents such as the requirements specification and the functional specification is reinforced by the requirement to identify flaws as soon as possible. Static testing methods like inspections and walkthroughs are used for this.

    3. It introduces the concept of outlining test specifications and anticipated results before carrying out actual tests. For instance, rather than being performed against some criteria conceived once the acceptance stage has been reached, the acceptance tests are carried out against a specification of requirements.

    4. The V model offers a focus for outlining the testing requirements for each stage. The concept of entry and exit criteria aids in the definition of testing. Consequently, the model can be applied to specify the state that a deliverable must be in before entering and leaving each stage. Typically, the entry criteria for one stage are the exit criteria for the following The quality of the program code released by individual programmers is a concern in many organizations. Some programmers publish code that appears to be error-free, while others publish code that still contains a significant number of errors. The exit criteria for unit design and unit testing would address the issue of programmers releasing code with varying levels of robustness. Before writing any program code, programmers would need to specify their intended test cases using unit design. Coding couldn’t start until these test cases were approved by the proper manager. Second, before the program could move from the unit test stage to integration testing, the test cases would need to be completed successfully.

    5. Last but not least, the V model offers a foundation for identifying who is in charge of conducting the testing at each stage. Here are some typical responsibilities:

  • acceptance testing performed by users
  • system testing performed by system testers
  • integration testing performed by program team leaders
  • unit testing performed by programmers.
  • The V model is a great starting point for testing partitioning because it emphasizes that everyone involved in system development is accountable for quality assurance and testing.

    25. What do the terms “white box testing” and “black box testing” mean?

    A method of software testing called “white-box testing” (also known as “clear box testing,” “glass box testing,” “transparent box testing,” and “structural testing”) examines an application’s internal mechanisms rather than its functionality (i.e. e. black-box testing). In white-box testing, test cases are created using both programming knowledge and an internal perspective of the system. The tester selects inputs to test various code paths and determine the necessary outputs. This is analogous to testing nodes in a circuit, e. g. in-circuit testing (ICT).

    While white-box testing can be used during the software testing process at the unit, integration, and system levels, it is typically done at the unit level. When performing a system-level test, it can test the paths within a unit, the paths between units during integration, and the paths between subsystems. Although this approach to test design can find a lot of errors or issues, it might miss requirements that aren’t implemented or parts of the specification that aren’t implemented.

    Black-box testing is a type of software testing that focuses on an application’s functionality rather than its internal workings or structures (see white-box testing). All levels of software testing, including unit, integration, system, and acceptance testing, can use this test methodology. It typically dominates higher level testing, if not all of it, but it can also completely take over unit testing.

    White box testing refers to testing an application using programming or coding skills. That means the tester has to correct the code also.

    Testing an application “black box” means that the tester is not required to have coding or programming knowledge. He only looks at the GUI features and the application’s external functional behavior.

    Sl.No Black Box White Box
    1 Focuses on the functionality of the system Focuses on the structure (Program) of the system
    2 Techniques used are :

    • Equivalence partitioning
    • Boundary-value analysis
    • Error guessing
    • Race conditions
    • Cause-effect graphing
    • Syntax testing
    • State transition testing
    • Graph matrix
    Techniques used are:

    • Basis Path Testing
    • Flow Graph Notation
    • Control Structure Testing

    1. Condition Testing

    2. Data Flow testing

    • Loop Testing

    1. Simple Loops

    2. Nested Loops

    3. Concatenated Loops

    4. Unstructured Loops

    3 Tester can be non technical Tester should be technical
    4 Helps to identify the vagueness and contradiction in functional specifications Helps to identify the logical and coding issues.

    26. What are the types of testings?

    27. The Difference between Bit Rate and Baud Rate?

    The distinction between bit rate and baud rate is intricate and interconnected. Both are dependent and inter-related. However, the most straightforward definition of a bit rate is the number of data bits transmitted per second. The baud rate is the frequency at which a communications channel’s signal changes per second.

    The number of data bits—0s and 1s—transmitted in a communication channel per second is known as the bit rate. The abbreviation “bps” refers to the number of zeros or ones that can be transmitted in a second when it is stated as 2400 bits per second. “Each individual character (such as a letter or a number), also known as a byte, is made up of several bits.

    The number of times a signal in a communications channel changes state or fluctuates is known as the baud rate. For instance, a channel can change states up to 2400 times per second at a 2400 baud rate. When something is said to “change state,” it can alternate between 0 and 1 or 1 and 0 up to X (in this case, 2400) times per second. Additionally, it refers to the connection’s actual state, such as its voltage, frequency, or phase level.

    The primary distinction between the two is that depending on the modulation technique, one change of state can transmit one bit or slightly more or less than one bit. The relationship between bit rate (bps) and baud rate (baud per second) is as follows:

    bps is equal to baud per second times the amount of bits per baud.

    The modulation technique determines the number of bit per baud. Here are two examples:

    Each baud transmits one bit when frequency shift keying (FSK) is used as the transmission method. To send a bit, only one change in state is necessary. The bps rate of the modem is therefore equal to the baud rate. Phase modulation is a type of modulation that transmits four bits per baud when a baud rate of 2400 is used. So:

    2400 baud x 4 bits per baud = 9600 bps

    Such modems are capable of 9600 bps operation.

    3)Difference between flash and EEprom

  • Both flash and EEPROM are digital storage methods used by computers and other devices. Both are non-volatile ROM technologies to which you can write and from which you can erase multiple times.
  • The primary difference between flash and EEPROM is the way they erase data. While EEPROM destroys the individual bytes of memory used to store data, flash devices can only erase memory in larger blocks. This makes flash devices faster at rewriting, as they can affect large portions of memory at once. Since a rewrite may affect unused blocks of data, it also adds unnecessarily to usage of the device, shortening its lifespan in comparison with EEPROM.
  • Flash storage is commonly used in USB memory drives and solid state hard drives. EEPROM is used in a variety of devices, from programmable VCRs to CD players.
  • 28. Can structures be passed to the functions by value?

    Ans: yes structures can be passed by value. But unnecessary memory wastage.

     #include  main() { Func1(); Func2(); }
     /************************************* * * Function declarations (prototypes). * *************************************/ /* Func1 is only visable to functions in this file. */ static void Func1(void); /* Func2 is visable to all functions. */ void Func2(void); /************************************* * * Function definitions * *************************************/ void Func1(void) { puts("Func1 called"); } /*************************************/ void Func2(void) { puts("Func2 called"); }

    31. Difference between declaration, definition & initialization?

    Ans: A declaration gives the compiler a name, or an identifier. The statement instructs the compiler that “This function or this variable exists somewhere, and this is how it should look.” ”.

    In contrast, a definition specifies: “Make this variable here” or “Make this function here.” ” It allocates storage for the name. This interpretation holds true whether you’re referring to a variable or a function; in both cases, the compiler allots space at the definition stage.

    This initialization establishes that this is a definition and not a declaration. extern const int x = 1;

    This C++ declaration indicates that the definition is found elsewhere. extern const int x;

    32. What distinguishes pass by reference in C from pass by value by reference in C? Pass By Reference: In pass by reference, a variable’s address is passed to a function. The same memory location is used for both variables, so any changes made to the formal parameter will also affect the actual parameters. When you need to return multiple values, (Formal and Actual) is useful Pass By Value: – This method passes the variable’s value. Changes made to formal will not affect the actual parameters. – Different memory locations will be created for both variables. – In this case, a temporary variable will be created in the function stack without affecting the primary variable.


    What is hardware in the loop testing?

    Hardware-in-the-loop (HIL) is a term used in the automotive industry to describe a technique for testing and validating complicated software systems on test benches with specialized hardware that takes data inputs from real-world gadgets like radars and cameras.

    What is the difference between HIL and SIL?

    SIL: Software In the Loop. In order to correct issues with system functionality (code generation validation), unit tests are run on the code that will be incorporated into a calculator. HIL: Hardware In the Loop.

    What is the difference between MIL SIL PIL and HIL test?

    Verification (simulation) and validation (testing) are key elements of MBSE. To guarantee a solid and trustworthy outcome, the MBSE process includes specific instances of model in the loop (MIL), software in the loop (SIL), processor in the loop (PIL), and hardware in the loop (HIL) simulation and testing.

    What is processor in the loop?

    A processor-in-the-loop (PIL) simulation downloads and executes object code on your target hardware after cross-compiling generated source code. You can determine whether your model and the generated code are numerically equivalent by comparing the results of normal and PIL simulations.

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