Steps involved in converting the C source code to executable:
Now, we are going to see how to convert a C source code to executable and also about the steps involved in compiling a C program. C program compilation involves 4 steps which are as follows.
Here, we have two c files main.c and add.c.
Below is the source code for main.c
Now, we are going to see how to convert a C source code to executable and also about the steps involved in compiling a C program. C program compilation involves 4 steps which are as follows.
- Converting C source code to Preprocessed code
- Converting preprocessed code to assembly code
- Converting assembly code to object code
- Converting object code to binary
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| Steps involved in compiling a C Program |
Here, we have two c files main.c and add.c.
Below is the source code for main.c
#define NUM1 10
#define NUM2 20
int main() {
int a, b;
a = NUM1;
b = NUM2;
add(a, b);
return 0;
}
Below is the source code for add.c
#include <stdio.h>
#define STRING "Result:"
void add(int a, int b) {
int c;
c = a + b;
printf("%s %d\n", STRING, c);
return;
}
As I mentioned earlier, compiling a C program involves 4 phases. They are preprossesing, compilation, assembly and linking. We don't need to go through all the above four steps one by one to get final executable for the given source code. Because, compiler can do all the above operation implicitly and provide us the final binary or executable in one single command(see below).
gcc <file_name.c> -o <execuable_name>
Please note that the above command itself can get us the executable for the given source code. Just to understand whole compilation process, we are going to go through all the steps involved in compilation one by one.
What is preprocessor?
It converts the C source code to preprocessed code or expanded c source code. In other words, its a program which runs on our C source code before compilation process. And it provides facility to handle named constant, macros and file inclusion. Preprocessor begins with the preprocessor directive hash symbol(#). During preprocessing the following operation would take place.
- macro substitution
- File inclusion
- Conditional compilation
How to convert C source code to preprocessed code?
We can use -E option to perform preprocessing alone. The below command converts the C source code to preprocessed code. The -o option is used to redirect the preprocessed code to the given file.
gcc -E main.c -o main.i -> redirects the preprocessed code to the file main.i
gcc -E add.c -o add.i -> redirects the preprocessed code to the file add.i
jp@jp-VirtualBox:~/$ gcc -E main.c -o main.i
jp@jp-VirtualBox:~/$ gcc -E add.c -o add.i
jp@jp-VirtualBox:~/$ ls
add.c add.i main.c main.i
jp@jp-VirtualBox:~/$ gcc -E add.c -o add.i
jp@jp-VirtualBox:~/$ ls
add.c add.i main.c main.i
Below is the preprocessed code(main.i) of main.c
# 1 "main.c"
# 1 "<built-in>"
# 1 "<command-line>"
# 1 "main.c"
int main() {
int a, b;
a = 10;
b = 20;
add(a, b);
return 0;
}
Below is the preprocessed code(add.i) of add.c
# 936 "/usr/include/stdio.h" 3 4
# 2 "add.c" 2
void add(int a, int b) {
int c;
c = a + b;
printf("%s %d\n", "Result:", c);
return;
}
Note: In the above preprocessed source codes, you could find that the macros are replaced by values.
What is a compiler?
Compiler converts the preprocessed(or C) source code to assembly language code.
How to convert preprocessed source code to assembly language code?
The -S option tells the compiler to generate assembly language code alone. The below command converts the preprocessed source code to assembly language code. The -o option is used to redirect the assembly language code to the given file.
gcc -S main.i -o main.s -> redirects the assembly code for main.i to the file main.s
gcc -S add.i -o add.s -> redirects the assembly code for add.i to the file add.s
jp@jp-VirtualBox:~/$ gcc -S main.i -o main.s
jp@jp-VirtualBox:~/$ gcc -S add.i -o add.s
jp@jp-VirtualBox:~/$ ls
add.c add.i add.s main.c main.i main.s
jp@jp-VirtualBox:~/$ gcc -S add.i -o add.s
jp@jp-VirtualBox:~/$ ls
add.c add.i add.s main.c main.i main.s
Below is the assembly code(main.s) for main.i
.file "main.c"
.text
.globl main
.type main, @function
main:
pushl %ebp
movl %esp, %ebp
andl $-16, %esp
subl $32, %esp
movl $10, 28(%esp)
movl $20, 24(%esp)
movl 24(%esp), %eax
movl %eax, 4(%esp)
movl 28(%esp), %eax
movl %eax, (%esp)
call add
movl $0, %eax
leave
ret
.size main, .-main
.ident "GCC: (Ubuntu/Linaro 4.4.4-14ubuntu5.1) 4.4.5"
.section .note.GNU-stack,"",@progbits
Below is the assembly code(add.s) for add.i
.file "add.c"
.section .rodata
.LC0:
.string "%s %d\n"
.LC1:
.string "Result:"
.text
.globl add
.type add, @function
add:
pushl %ebp
movl %esp, %ebp
subl $40, %esp
movl 12(%ebp), %eax
movl 8(%ebp), %edx
leal (%edx,%eax), %eax
movl %eax, -12(%ebp)
movl $.LC0, %eax
movl -12(%ebp), %edx
movl %edx, 8(%esp)
movl $.LC1, 4(%esp)
movl %eax, (%esp)
call printf
leave
ret
.size add, .-add
.ident "GCC: (Ubuntu/Linaro 4.4.4-14ubuntu5.1) 4.4.5"
.section .note.GNU-stack,"",@progbits
What is assembler?
Assembler converts the assembly language code to object code.
How to convert assembly language code to object code?
The -c option tells the compiler to generate object code(machine codes). The below command converts the assembly language code to object codes. The -o option is used to redirect the object code to the given file.
gcc -c main.s -o main.o -> redirects the object code for main.s to the file main.o
gcc -c add.s -o add.o -> redirects the object code for add.s to the file add.o
jp@jp-VirtualBox:~/$ gcc -c add.s -o add.o
jp@jp-VirtualBox:~/$ ls
add.c add.i add.o add.s main.c main.i main.o main.s
jp@jp-VirtualBox:~/$ ls
add.c add.i add.o add.s main.c main.i main.o main.s
What is linker?
Linker links one or more object files and generates a single executable.
How to convert objects codes to an executable?
The below command links the given object files and generates an executable file. The -o option is used to mention the name of the executable file.
gcc main.o add.o -o output -> output is the executable file
jp@jp-VirtualBox:~/$ gcc main.o add.o -o output
jp@jp-VirtualBox:~/$ ls
add.c add.i add.o add.s main.c main.i main.o main.s output
jp@jp-VirtualBox:~/$ ./output
Result: 30
jp@jp-VirtualBox:~/$ ls
add.c add.i add.o add.s main.c main.i main.o main.s output
jp@jp-VirtualBox:~/$ ./output
Result: 30
./output - "./" indicates current directory. And "output" is the name of the executable.
So, our program prints the sum value of two numbers 10 and 20.
