Question

Question 1: 1). Mr. Mahmood is a diploma student from engineering department. He is studying "computer programming for engineering” course. He wants to know how the 'C' program is converted into machine code and the corresponding program which is used to convert the 'C' program into machine code line by line and the program which is used to convert the whole program into machine code at once. Provide the required information in detail to Mr. Mahmood. -2 marks Answer:

Answer 1

Answer:-

The procedure for line by conversion of a C program is shown below.

• Process of compilation does the conversion of C program into object file which the is passed on to the assembler which does job of converting to machine code.
• Which will be hex values or binary values.

Example:-

1. Let's compile the c program for Fibonacci series. line by line:-

#include <stdio.h> int main(void) { int x, y, z; while (1) { X = 0; y = 1; do { printf("%d\n", x); z = x + y; X = y; y = z; } while (x < 255); > }

• Now if we try to see the working, what it simply does is,
• Choses first two numbers, adds them and stores in third variables, prints the values and stores back third value to second variable and value from second variable to first.

NOW Let's decode:

OK8 x % cat fib.c #include <stdio.h> int main(void) { int x, y, z; 1 7 while (1) { X = 0; y = 1; do { 2 printf("%d\n", x); fib: (_TEXT,_text) section main: eeeee00100000f20 pusha 0000000100000f21 mova 0000000100ee0f24 subq 0000000100000f28 movi 0000000100000f2f movi ee00000100000f36 movl e000000100000f3d leaq 0000000100000f44 movl 0000000100000f47 movb 0000000100000f49 calla 0000000100000f4e movl 0000000100000f51 addi 0000000100000f54 movi 0000000100000f57 mov] 0000000100000f5a. movi 0000000100000f5d movi 0000000100000f60 movi 0000000100000f63 movl 0000000100000f66 cmpl 0000000100000f6d ji AAAAAAA199000f73 imp %rbp Ox 10 근 %rsp, %rbp $0x20, %rsp $0x0, -0x4(%rbp) $0x0, -0x8(%rbp) XO $0x1, -exc(%rbp) Yal 0x56(%rip), %rdi -Ox8(%rbp), %esi 2 printt X $exo, %al Ox1000e0f78 -Ox8(%rbp), %esi X-ES/ -Oxc(%rbp), %esi 2=xty %esi, -0x10(%rbp) »Z -Oxc(%rbp), %esi Yey 4 %esi, -@x8(%rbp) -Ox10(%rbp), %esi Y=z 5 %esi, -exc(%rbp) %eax, -ex14(%rbp) 6 $0xff, 0x8(%rbp) 0x100000f3d Ox100000f2f 7 3 d Y-7251 3 Z = X + y; 4 X = y; 5 y = z; 6 } while (x < 255); }

• On the left you see the converted hex code and aside of it is the respected Assembly code.
• SO first it has to be converted to assembly then to hex and binary.
• The numberings you see are the sequences of the C program ( on right) being converted to assembly.
• 1. The vales of X = 0 and Y =0 is set using move instruction. Pushed in stack.
• The Values of $0x0 are the memory location on which values will be temporarily stored. On RAM.
• 2. Printing of Value from X is done.
• Then comes the summation and assignment part.
• Then at the end shuffle of values takes place.
• and then comes the looping part ( In assembly loops comes last as the condition is to be checked)
• That's how control of execution of a c program actually works.