Embedded engineer Interview Questions
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Embedded Engineer interview questions shared by candidates
Given the following struct definition in C, what would sizeof(myStruct) return? typedef struct { short a; long b; char c; } myStruct;
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For the 64 bit system, typedef struct { short a; //2 bytes long b; //8 bytes char c; //1 bytes } As alignment is due to largest type : 2 bytes(short) + 6bytes(padding to achieve 8 byte boundry) + 8bytes(long) + 1byte(char) + 7 bytes(extra padding to achieve 8 byte boundry) = 24bytes Less
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if we using 32 bit the answer defiantly will be 7+1 for padding -> 8 bytes if we using 64 bit the answer will be 17+7 for padding =24 bytes Less
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(Hint: you have to keep in mind the rules of byte padding since #pragma pack(1) is not called) Less
Questions on C and details of the previous projects carried out
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What type of questions brother Like moderate or easy
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Moderate level
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Can you share the Questions, from which topic they asked in c
reverse bits of an integer that is a power of 2 , keeping complexity in mind.
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if the bits that need to be reversed are just the ones to the right of the 1 so all you need to do is: return n -1; 1000 -> 0111 1 -> 0 10 -> 1 10000000 -> 1111111 Less
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int reverse(int n) { int msb=Math.log(n)/Math.log(2); return Math.pow(2, 32-msb); } Less
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rev_n = n ^ (~(n*0));
What are different types of interrupts?
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External &Internal Interrupt
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1- External Interrupts: These types of interrupts generally come from external input / output devices which are connected externally to the processor. They are generally independent and oblivious of any programming that is currently running on the processor. 2- Internal Interrupts: They are also known as traps and their causes could be due to some illegal operation or the erroneous use of data. Instead of being triggered by an external event they are usually triggered due to any exception that has been caused by the program itself. Some of the causes of these types of interrupts can be due to attempting a division by zero or an invalid opcode etc. 3- Software interrupts: These types if interrupts can occur only during the execution of an instruction. They can be used by a programmer to cause interrupts if need be. The primary purpose of such interrupts is to switch from user mode to supervisor mode. Less
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Cyclic and Eventual. (this answer is not the best, because it describes when an interrupt occurs. It should be that there are software interrupts and hardware interrupts). Less
Q1- Use of Volatile keyword?
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volatile your a. ., does that count I used it
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wrong answer, adjective, adjective, should be removed, constructive answer: exclude from optimising Less
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"volatile" keyword in C/C++ is used for any variable whose value can change anytime unexpectedly for example variables used in ISR, Memory mapped peripheral registers and variables used by multiple threads in RTOS based applications. Less
why should i hire you
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i too attened the interview on the same date . Now i have been called for managerial round. what questions thy asked u in managerial round. can u please tell Less
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they asked puzzles based on real time scenario.they will check your attitude and decision making skills. Less
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when you attended the interview .is it on 30th may 2015 ?
Is the given number a bitwise palindrome?
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I took it as without proceeding zeroes. Also, you can cut the cycles in half by only going halfway through the bits. uint8_t isPalindrome(uint input) { uint msb = log2(input); for (int i = 0; i < msb / 2; i++) { uint8_t left = ((1 << (msb - i)) & input) != 0 ? 1 : 0; uint8_t right = ((1 << i) & input) != 0 ? 1 : 0; if (left != right) { return 0; } } return 1; } Less
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#include int main(void) { //1001001 int num = 0x49246; int count = 0; int a = num; while(a > 0){ printf("%d\n", a & 1); a >>= 1; count += 1; } int left = 0, right = count - 1; while (left > left) & 1) != ((num >> right) & 1)){ printf("No"); return 1; } left += 1; right -= 1; } printf("Yes"); return 0; } Less
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boolean isPalindrome(int n) { for(int i=0; i<32; i++) { int temp1=n&(1< Less
will cache affect memory I/O register?
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cache affects memory I/O reg. When using peripherals you want memory accesses to happen in a certain order (mostly because of hardware constraints) so you want to avoid reordering. This is exactly what you would expect cache to do; temporarily store instructions and perform optimizations by grouping/reordering. You might get unexpected errors because of this. Less
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Depends if the I/O register is memory mapped, processor type and a host of other things.. Less
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this is related to cache coherency.
Accurately (this is the catch) read a 64bit register value using a method that can read only 32bit at a time
4 Answers↳
uint32_t read_context; memcpy(&read_context, (void *)(REGISTER_BASE_ADDR), 4); printf("Register MSW: %x \t", read_context, 0, 0, 0, 0, 0); memcpy(&read_context, (void *)(REGISTER_BASE_ADDR + 4), 4); printf("Register LSW: %x \t" , read_context, 0, 0, 0, 0, 0); Less
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uint64_t result = (uint64_t) read32(ptr1) | read32(ptr1+4)
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#include #include void register_read(void* reg,void *result); int main() { //printf("Hello World"); uint64_t reg=0xAAAAAAAAFFFFFFFF,result; register_read((void*)®,(void*)&result); printf("%lX",result); return 0; } void register_read(void* reg,void *result){ uint32_t *ptr1,*ptr2; ptr1=(uint32_t *)(reg); ptr2=(uint32_t *)(result); for(int i=0;i<2;i++){ *ptr2=*ptr1; ptr1++; ptr2++; } Less
A brain teaser question where we have to find out 45 minutes with the help of two ropes. Given that one rope burns completely in 1 Hr and the rate or burning is not consistent.
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Burn first rope from both ends, and second rope from one end only. When First has completely burned, 30 mins will have passed and second rope will have 30 mins left on it. Now burn second rope, which has burned for 30 mins already, from both ends, this will burn a 30 minute rope at twice speed, making it complete in 15 mins. This will be 45 minutes total. Less
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I've faced same question in ASSIA interview
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I assume that both ropes have the same non consistency. If you burn from one end it takes 1H. If you burn the first rope from both ends it takes 1/2 H. Immediately after the first rope burnt, burn the second rope from one end and the middle point that fires reached each other in the first rope. To get 1/4 H, burn it from both ends and the point that in the first rope the fires got together. Less