在C程序中while（i = 20）或任何数字的结果是什么？

Question

Came across this question somewhere-

What is the output of the following program

int main(void) {
    int  i = 10 ; 
    while ( i = 20 )  
    printf ( "\nA computer buff!" ) ;
    return 0;
}

Now I know that every non-zero number is treated as true in C but I haven't been able to figure out the result of using an assignment operator inside while. When I run the code "A computer buff" got printed unknown number of times in a loop. I know it's a stupid question but question nonetheless.

Answer 1

That should be an infinite loop, because i = 20 will yield 20 as a result of the assignment expression.

From the docs

An assignment operation assigns the value of the right-hand operand to the storage location named by the left-hand operand. Therefore, the left-hand operand of an assignment operation must be a modifiable l-value. After the assignment, an assignment expression has the value of the left operand but is not an l-value.

Answer 2

In your code

 while ( i = 20 )

is essentially assigning 20 to i and then taking the i as the conditional check for while loop. As there in no break ing condition inside the loop body, it is effectively an infinite loop.

Answer 3

You can easily check it on *nix by using the -S option in gcc. I took this file:

#include <stdio.h>

main( ){  
   int  i = 10 ; 
   while ( i == 20 )  
   printf ( "\nA computer buff!" );
}

with gcc -S test.c and it gave me out this assambly:

    .file   "c.c"
    .section    .rodata
.LC0:
    .string "\nA computer buff!"
    .text
    .globl  main
    .type   main, @function
main:
.LFB0:
    .cfi_startproc
    pushq   %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset 6, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register 6
    subq    $16, %rsp
    movl    $10, -4(%rbp)
    nop
.L2:
    movl    $20, -4(%rbp)
    movl    $.LC0, %edi
    movl    $0, %eax
    call    printf
    jmp .L2
    .cfi_endproc
.LFE0:
    .size   main, .-main
    .ident  "GCC: (Ubuntu 4.8.4-2ubuntu1~14.04) 4.8.4"
    .section    .note.GNU-stack,"",@progbits

You can see at jmp .L2 that this is the only branch, responsible for the loop, which is uncondicional. movl $20, -4(%rbp) is the assignment of 20 to i.

Answer 4

while(1) is infinite loop so its execute number of time. If you want to move out from the loop after the first iteration then use break statement in loop.