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      Junior Compiler Software Engineer Interview

      29 Mar 2018
      Anonymous interview candidate
      Herzliya
      No offer
      Neutral experience
      Difficult interview

      Application

      I applied online. The process took 1 day. I interviewed at CEVA (Herzliya) in Mar 2018

      Interview

      3 questions in the interview: 1. Implement the next function in C (value can be also negative): char * itoa (int value, char * str); 2. I don't remember the question. 3. Assume your assembly language includes ONLY the following instructions: 'inc REG': increments a given register by one. 'dec REG': decrement a given register by one. 'jnz LABEL': jumps to a given LABEL if the previous instruction's result was not zero. 'HALT': stops running. Task: A and B registers hold non-negative values. The program should calculate the value of |A-B| and locate the result in C. In addition, the language holds registers C,D,...,Z, which you can assume are initialized at program start to zero.

      Interview questions [2]

      Question 1

      Implement the next function in C (value can be also negative): char * itoa (int value, char * str);
      1 Answer

      Question 2

      Assume your assembly language includes ONLY the following instructions: 'inc REG': increments a given register by one. 'dec REG': decrement a given register by one. 'jnz LABEL': jumps to a given LABEL if the previous instruction's result was not zero. 'HALT': stops running. Task: A and B registers hold non-negative values. The program should calculate the value of |A-B| and locate the result in C. In addition, the language holds registers C,D,...,Z, which you can assume are initialized at program start to zero.
      2 Answers

      Other Junior Compiler Software Engineer interview reviews for CEVA

      Junior Compiler Software Engineer Interview

      12 Apr 2021
      Anonymous employee
      Tel Aviv-Yafo
      Accepted offer
      Positive experience
      Easy interview

      Application

      I applied online. I interviewed at CEVA (Tel Aviv-Yafo) in Jan 2021

      Interview

      extern short const_arr[N] long func(short x) { static short array[N]; long res = 0; int i; for (i = N-1; i > 0;i--) array[i] = array[i - 1]; // 1 arithmetic operation + 2 operations R/W = 11 cyc * (N-1) array[0] = x; // 1 R/W = 5 cyc for (i = 0; i < N; i++) res += (long)array[i] * const_arr[i]; //2 R/W + 2 arithmetic op = 12 * N cyc return res; }
 arr = {1,2,3,4} const_arr = {8,9,10,11} func(7) arr = {1,2,3,7} res = 2. Calculate the overall running time of the above C code as a function of the array's length (n=N) . Assume that: A. Arithmetic operations (such as +,-,*,=,+=) cost one machine cycle. B. Reading/Writing from an array element costs 5 machine cycles. C. Loops have a zero overhead e.g.: for (j = 0; j < n; j++) arr[j] = a + b; The cost of the loop is 6*n machine cycles (5 writing arr[j] ,1 calculating a+b). "j = 0", "j < n", "j++" are not taken into consideration when calculating cycle count.
 *************************************** 3. Lets assume that we add the following 2 new operators to the C language: 1) inc_mod(i, n); i++; if (i >= n) i = 0; 2) dec_mod(i, n); i--; if (i < 0) i = n - 1; How can the function func() be optimized SIGNIFICANTLY in terms of run time. Assume that the function func() will be called much more than N times (thus optimizing the first multiplications of zero array elements, is not worthwhile).


      Interview questions [1]

      Question 1

      Primitive Assembley Question: ============================= Assume your assembley language includes ONLY the following instructions: - 'inc REG': increments a given register by one. - 'dec REG': decrements a given register by one. - 'jnz LABEL': jumps to a given LABEL if the previous instruction's result was not zero. - 'HALT': stops running. Task: A and B registers hold non-negative values. The program should calculate the value of |A-B| and locate the result in C. In addition the language holds registers C, D, ..., Z, which you can assume are initialized at program start to zero.
      1 Answer
      1

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