Embedded software engineers develop and maintain embedded systems, which may require advanced software skills and problem solving. In an embedded software engineer interview, the interviewer may ask questions that test your coding skills and experience. Expect to also discuss how you work with others, as the role may function within a team.
Here are three top embedded software engineer interview questions and how to answer them:
How to answer: Prove your competency by delivering a concise but thorough answer. Explain that an embedded system is a computer that lives within a bigger machine and operates in a unique way to perform a particular task. Consider offering some examples, like a car, a kitchen appliance, or an MP3 player.
How to answer: Embedded software engineers are responsible for building complex functional systems, so it's beneficial to highlight your technical skills. The interviewer may want to hear you mention basic programming languages like C and C++. You might mention which languages you prefer or which you would like to gain more experience in, demonstrating passion and an ongoing willingness to learn.
How to answer: You may work with others towards a common goal, so teamwork is vital. Consider sharing an anecdote about a time when you collaborated successfully to resolve an issue with a product. Focus on your ability to manage stress, clearly communicate complex issues, and listen to others.
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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
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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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rev_n = n ^ (~(n*0));
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uint flipBits(uint input) { uint bits = sizeof(input) * 8; uint msb = log2(input); return 1 << (bits - msb); } Less
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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
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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
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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 ?
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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
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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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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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may be not..
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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
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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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#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
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uint64_t result = (uint64_t) read32(ptr1) | read32(ptr1+4)
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It returns an integer - the status of the operation
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The key in these questions is to cover the fundamentals, and be ready for the back-and-forth with the interviewer. Might be worth doing a mock interview with one of the Fitbit or ex-Fitbit Embedded Software Engineer experts on Prepfully? They give real-world practice and guidance, which is pretty helpful. prepfully.com/practice-interviews Less
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A pointer to dest so that memcpy could be used as an argument in another function. Less