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      Entrevista para Junior Compiler Software Engineer

      29 mar 2018
      Candidato de entrevista anónimo
      Herzliya

      Otras evaluaciones sobre las entrevistas para el cargo de Junior Compiler Software Engineer en CEVA

      Entrevista para Junior Compiler Software Engineer

      12 abr 2021
      Empleado anónimo
      Tel Aviv
      Oferta aceptada
      Sin ofertas
      Experiencia neutra
      Entrevista difícil

      Solicitud

      Me postulé en línea. El proceso tomó 1 día. Acudí a una entrevista en CEVA (Herzliya) en mar 2018

      Entrevista

      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.

      Preguntas de entrevista [2]

      Pregunta 1

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

      Pregunta 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 respuestas
      Experiencia positiva
      Entrevista fácil

      Solicitud

      Me postulé en línea. Acudí a una entrevista en CEVA (Tel Aviv) en ene 2021

      Entrevista

      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).


      Preguntas de entrevista [1]

      Pregunta 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 respuesta
      1