Why the output is 2 2 2 if I put static

Question

1.00/5 (1 vote)

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C

int main ()
{
static int i=1;

printf("%d %d %d",i++,i,++i);

return 0;
}

What I have tried:

I have tried the same query using static

Posted 15-Sep-23 20:54pm

Raja Ram 2023

Updated 16-Sep-23 1:56am

merano99

v2

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Comments

Richard MacCutchan 17-Sep-23 3:37am

I just tried this in pure C and get the result 2 3 3, whether i is static or not. The take home message being: do not use expressions like this.

0x01AA 17-Sep-23 10:25am

And I don't see a logic for that result :-)

Richard MacCutchan 17-Sep-23 10:37am

In a C++ program it prints as 2 2 2. So there is no (real) logic in either case. The only way to see what is happening is by generating an assembler listing. I did discover a bug in the increment operator in the C compiler earlier this year, so this could be a manifestation of the same or a similar problem.

0x01AA 17-Sep-23 10:45am

In c++ there is a 'logic'. The logic is to evaluate parameters from right to left. But of course, I would not rely on, that every compiler does it the same way.
Therefore, your statement 'The take home message being: do not use expressions like this.' is very ok but unfortunately I can't give you a five here for a comment ;)

Richard MacCutchan 17-Sep-23 11:14am

See Solution 5.

0x01AA 17-Sep-23 11:26am

Of course a 5

Richard MacCutchan 17-Sep-23 11:28am

Thank you, an interesting exercise.

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merano99 · Answer 1 · 2023-09-16T02:03:00

Solution 3

Obviously the pre-increment is used first here. As 0x01AA has already noted, the result would otherwise be different. This kind of implementation is unreliable, but here the code and result make the question clear.

Why should the keyword static change the result here?

Posted 16-Sep-23 2:03am

merano99

Updated 16-Sep-23 2:05am

v2

Comments

0x01AA 16-Sep-23 9:24am

My 5. And yes, 'static' does not (should never) have any effect.

Richard MacCutchan · Answer 2 · 2023-09-17T05:13:00

Some information that may be of interest ... (or not, as the case may be)

1. The assembler generated by the C compiler

ASM

; 10   :     static int i=1;
; 11   : 
; 12   :     printf("%d %d %d",i++,i,++i);

	mov	eax, DWORD PTR ?i@?1??ctest@@9@9 ; eax = 1
	add	eax, 1                           ; eax = 2
	mov	DWORD PTR ?i@?1??ctest@@9@9, eax ; i = 2
	mov	ecx, DWORD PTR ?i@?1??ctest@@9@9 ; ecx = 2
	mov	DWORD PTR tv69[ebp], ecx         ; tv69 = 2
	mov	edx, DWORD PTR ?i@?1??ctest@@9@9 ; edx = 2
	add	edx, 1                           ; edx = 3
	mov	DWORD PTR ?i@?1??ctest@@9@9, edx ; i = 3
	mov	eax, DWORD PTR ?i@?1??ctest@@9@9 ; eax = 3
	push	eax                          ; push 3 -----> third value
	mov	ecx, DWORD PTR ?i@?1??ctest@@9@9 ; ecx = 3
	push	ecx                          ; push 3 -----> second value
	mov	edx, DWORD PTR tv69[ebp]         ; edx = 2
	push	edx                          ; push 2 -----> first value
	push	OFFSET $SG10501              ; push format string
	call	_printf

2. The assembler generated by the C++ compiler

ASM

; 18   :     static int i=1;
; 19   : 
; 20   :     printf("%d %d %d",i++,i,++i);

	mov	eax, DWORD PTR ?i@?1??CppTest@@YAHXZ@4HA ; eax = 1
	add	eax, 1                                   ; eax = 2
	mov	DWORD PTR ?i@?1??CppTest@@YAHXZ@4HA, eax ; i = 2
	mov	ecx, DWORD PTR ?i@?1??CppTest@@YAHXZ@4HA ; ecx = 2
	mov	DWORD PTR tv75[ebp], ecx                 ; tv75 = 2
	mov	edx, DWORD PTR ?i@?1??CppTest@@YAHXZ@4HA ; edx = 2
	mov	DWORD PTR tv73[ebp], edx                 ; tv73 = 2
	mov	eax, DWORD PTR ?i@?1??CppTest@@YAHXZ@4HA ; eax = 2
	mov	DWORD PTR tv69[ebp], eax                 ; tv69 = 2
	mov	ecx, DWORD PTR ?i@?1??CppTest@@YAHXZ@4HA ; ecx = 2
	add	ecx, 1                                   ; ecx = 3
	mov	DWORD PTR ?i@?1??CppTest@@YAHXZ@4HA, ecx ; i = 3
	mov	edx, DWORD PTR tv75[ebp]                 ; edx = 2
	push	edx                                  ; push 2 -----> third value
	mov	eax, DWORD PTR tv73[ebp]                 ; eax = 2
	push	eax                                  ; push 2 -----> second value
	mov	ecx, DWORD PTR tv69[ebp]                 ; ecx = 2
	push	ecx                                  ; push 2 -----> first value
	push	OFFSET $SG64299                      ; push format string
	call	_printf

Interestingly, if I remove the "-std:c++latest" option from the build
of the .cpp version, I get the result "2 3 3".

Mike Hankey · Answer 3 · 2023-09-15T21:22:00

Solution 1

The printf evaluates/converts the parameter from left to right;
since the i++ is a post increment it is passed.
The i, nothing needs to be done.
Then the ++i is a pre increment so i is incremented then the printf prints 222.

Posted 15-Sep-23 21:22pm

Mike Hankey

Comments

0x01AA 16-Sep-23 7:01am

You mean from right to left, right?
And I wouldn't rely on all compilers doing it this way. Altought it seems most do it ;)

OriginalGriff · Answer 4 · 2023-09-15T21:31:00

Solution 2

Because prefix and postfix operators don't work like you think: Why Does x = ++x + x++ Give Me the Wrong Answer?[^]

Posted 15-Sep-23 21:31pm

OriginalGriff

Patrice T · Answer 5 · 2023-09-16T19:19:00

Quote:
Why the output is 2 2 2 if I put static

You are in grey zone. This code is unpredictable and depends on compiler details and on optimizations, it can even change on different versions.
Never use a variable in a line of code with multiple increment/decrement of the variable. All you can predict is the value on next line of code.

Why the output is 2 2 2 if I put static

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