Does while(1){} do the same as while(1);

Question

I just want to be sure that this C code:

while(flag==true)
{
} 
foo();

does the same as this:

while(flag==true);

foo();

Have you tested it? Do you have ***any*** reason to believe it might be different at all? — abelenky, Mar 01 '17 at 14:39
And, don't use the second style, unless you just want the code to be look "cool" so that less people would understand, and introduce bugs in future. — Eric, Mar 01 '17 at 14:39
You know... The question is clear and well formatted, and yet I can't shake the feeling it should be off-topic for some reason... — StoryTeller - Unslander Monica, Mar 01 '17 at 14:39
Yes except you can place conditions in-between the braces of the first example, so you could add in a condition to change the state of the flag to false, thus exit the while loop. — Shaun Bebbers, Mar 01 '17 at 14:40
@StoryTeller: Maybe because title and contents don't match and can produce very different results. Qualifying `flag` as `volatile` can produce different behaviour, too. Which is the much more interesting (and relevant) question. — too honest for this site, Mar 01 '17 at 14:48
@StoryTeller I think this is a good question, maybe you just feel too much ... — Eric, Mar 01 '17 at 14:48
Some recommend `for(;;) { }` as an empty loop, since that removes the need for an expression. — unwind, Mar 01 '17 at 15:29

score 9 · Answer 1 · edited May 23 '17 at 12:09

9

; alone is a null statement in C.

In your case, {} or ; are syntactically needed, but they do the same: nothing

Related: Use of null statement in C

edited May 23 '17 at 12:09

Community

1
1

answered Mar 01 '17 at 14:40

Jean-François Fabre

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score 3 · Answer 2 · answered Mar 01 '17 at 14:47

3

In addition to the other answers: It's the same thing.

But I prefer this:

while (condition)
{
} 
foo();

over this:

while (condition);
foo();

because if you forget the semicolon after the while, your code will compile fine but it won't do what you expect:

while(condition)  // ; forgotten here
foo();

will actually be equivalent of:

while(condition)
{
  foo();
}

answered Mar 01 '17 at 14:47

Jabberwocky

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FYI: Recent gcc warns about this. I think it is the same as with Yoda-conditions: once they had their use, but now they just make the code less clear. You don't answer the question, but just state your personal opinion. – too honest for this site Mar 01 '17 at 14:52
@Olaf Using braces instead of a semi colon at the end is indeed good practice. Alternatively `while (condition) < new line > ;` is fine too. Harsh down vote, this is the best answer posted. The answer is also backed up by MISRA-C:2012 15.6 so this isn't just someone's subjective opinions. MISRA-C is in turn mostly based on scientific research, most notably by Les Hatton, who's work was published in the book _Safer C_. The semicolon issue is listed in said book, 2.7 Empirically Determined Misbehavior. This is a known source of bugs, proven by population studies of C code. – Lundin Mar 01 '17 at 15:28

score 2 · Answer 3 · answered Mar 01 '17 at 14:40

2

Yes, having an empty body of the loop is equivaled to just while(<some condition>);

answered Mar 01 '17 at 14:40

Gnqz

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score 2 · Answer 4 · answered Mar 01 '17 at 14:40

2

Yes. A ; following a control structure (e.g., while, for, etc.) that can be followed with a block is treated as if it was followed by an empty block.

answered Mar 01 '17 at 14:40

Mureinik

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msc · Answer 5 · 2017-03-01T14:53:45.223

2

Yes, because when put semicolon after while loop statement that indicate empty body and when the condition becomes false then it goes to the immediate next statement after that loop.

edited Mar 01 '17 at 14:53

answered Mar 01 '17 at 14:41

msc

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Suraj Jain · Answer 6 · 2017-03-01T17:03:49.497

Yes, they are same.

You Can Generate The assembly of the code and see for yourself that they produce the same assembly. (Using gcc filename.c -S -masm=intel -o ouputfilename)

#include<stdio.h>

int foo(void);
int main(){

  int flag;
  scanf("%d" , &flag);
  while(flag==1);
  foo(); 
}


int foo(void){
  int x = 2;
  return x*x;
}

.LC0:
    .ascii "%d\0"
    .text
    .globl  main
    .def    main;   .scl    2;  .type   32; .endef
    .seh_proc   main
main:
    push    rbp
    .seh_pushreg    rbp
    mov rbp, rsp
    .seh_setframe   rbp, 0
    sub rsp, 48
    .seh_stackalloc 48
    .seh_endprologue
    call    __main
    lea rax, -4[rbp]
    mov rdx, rax
    lea rcx, .LC0[rip]
    call    scanf
    nop
.L2:
    mov eax, DWORD PTR -4[rbp]
    cmp eax, 1
    je  .L2
    call    foo
    mov eax, 0
    add rsp, 48
    pop rbp
    ret
    .seh_endproc
    .globl  foo
    .def    foo;    .scl    2;  .type   32; .endef
    .seh_proc   foo
foo:
    push    rbp
    .seh_pushreg    rbp
    mov rbp, rsp
    .seh_setframe   rbp, 0
    sub rsp, 16
    .seh_stackalloc 16
    .seh_endprologue
    mov DWORD PTR -4[rbp], 2
    mov eax, DWORD PTR -4[rbp]
    imul    eax, DWORD PTR -4[rbp]
    add rsp, 16
    pop rbp
    ret
    .seh_endproc
    .ident  "GCC: (x86_64-posix-seh-rev1, Built by MinGW-W64 project) 6.3.0"
    .def    scanf;  .scl    2;  .type   32; .endef

And When I Changed while(flag == 1); to while(flag==1){} Assembly Code Generated is :

.LC0:
    .ascii "%d\0"
    .text
    .globl  main
    .def    main;   .scl    2;  .type   32; .endef
    .seh_proc   main
main:
    push    rbp
    .seh_pushreg    rbp
    mov rbp, rsp
    .seh_setframe   rbp, 0
    sub rsp, 48
    .seh_stackalloc 48
    .seh_endprologue
    call    __main
    lea rax, -4[rbp]
    mov rdx, rax
    lea rcx, .LC0[rip]
    call    scanf
    nop
.L2:
    mov eax, DWORD PTR -4[rbp]
    cmp eax, 1
    je  .L2
    call    foo
    mov eax, 0
    add rsp, 48
    pop rbp
    ret
    .seh_endproc
    .globl  foo
    .def    foo;    .scl    2;  .type   32; .endef
    .seh_proc   foo
foo:
    push    rbp
    .seh_pushreg    rbp
    mov rbp, rsp
    .seh_setframe   rbp, 0
    sub rsp, 16
    .seh_stackalloc 16
    .seh_endprologue
    mov DWORD PTR -4[rbp], 2
    mov eax, DWORD PTR -4[rbp]
    imul    eax, DWORD PTR -4[rbp]
    add rsp, 16
    pop rbp
    ret
    .seh_endproc
    .ident  "GCC: (x86_64-posix-seh-rev1, Built by MinGW-W64 project) 6.3.0"
    .def    scanf;  .scl    2;  .type   32; .endef

You can see that the relevant portion is same in both cases.

 //Below Portion is same in both cases. 
.L2:        
    mov eax, DWORD PTR -4[rbp]
    cmp eax, 1
    je  .L2
    call    foo
    mov eax, 0
    add rsp, 48
    pop rbp
    ret
    .seh_endproc
    .globl  foo
    .def    foo;    .scl    2;  .type   32; .endef
    .seh_proc   foo

good idea to disassemble both codes, not that it proves anything formally since it could well be implementation defined. — Jean-François Fabre, Mar 01 '17 at 17:17
@Jean-FrançoisFabre Sir, In that case how should i explain the OP? — Suraj Jain, Mar 01 '17 at 17:31

Does while(1){} do the same as while(1);

6 Answers6

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