Why 9 and 49 output ? 9 is correct but why k=49 ?

Question

1.80/5 (2 votes)

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C

#include <stdio.h>
#define PRODUCT(x) ((x)*(x))
int main()
{
    int i=3,j,k;
    j=PRODUCT(i++);
    k=PRODUCT(++i);
    printf("%d %d",j,k);
    return 0;
}

Posted 4-Nov-15 4:18am

Anupkumar_vj

Updated 4-Nov-15 4:22am

Thomas Daniels

v2

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Thomas Daniels 4-Nov-15 10:36am

Are you sure it outputs 9? When I ran it, I got 12. (I know why this happens, but if your first output is actually 9 then I have no idea because I would actually expect 12 here -- I'll answer if you can confirm that it's actually 12.)

Leo Chapiro 4-Nov-15 10:41am

Yes, take a look: http://codepad.org/2FyKYfmW

Thomas Daniels 4-Nov-15 10:46am

Huh, that's weird, because I get: http://ideone.com/FKOnAE

Leo Chapiro 4-Nov-15 10:58am

Yes, this is crazy! :) I have debugged with Visual Studio 2012 and got the same results (9 and 49).

Richard Deeming 4-Nov-15 11:43am

At a guess, one version is getting the value of x once, incrementing it twice as _duDE_ described, then returning 3 * 3.

The other is getting the value of x, incrementing it, getting the value of x again, incrementing it again, then returning 3 * 4.

It's probably one of those areas where the behaviour is "undefined", so if you feed the same code to three different compilers, you get five different results. :)

Anupkumar_vj 4-Nov-15 11:46am

No..In codeeblock I am getting 9 and 49.

PIEBALDconsult 4-Nov-15 10:45am

9 is certainly not correct.

Krunal Rohit 4-Nov-15 11:37am

It must be the compiler dependent outputs.

-KR

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Leo Chapiro · Accepted Answer · 2015-11-04T04:40:00

Solution 1

We can go through this code step-by-step:

int i=3,j,k;

The variables are declared, the "i" has value of 3.

j=PRODUCT(i++);

The macro PRODUCT is called, the value of "i" is first 3 and 3*3 = 9.
Now after this the value of "i" will be incremented twice because of (x)*(x)!
The value of "i" is now 5.

k=PRODUCT(++i);

The macro PRODUCT is called but before the value of "i" will be incremented twice because of (x)*(x) and is 7 and 7*7 = 49.

Posted 4-Nov-15 4:40am

Leo Chapiro

Comments

Krunal Rohit 4-Nov-15 11:45am

You're on fire _duDe_. 5!

-KR

Leo Chapiro 4-Nov-15 12:14pm

Thank you! :)

Anupkumar_vj 4-Nov-15 11:57am

i++*i++ means 3++*3++=3*3=9, i has incremented to 5. Now ++i*++i means ++5*++5=6*7=42 know ?

Dave Kreskowiak 4-Nov-15 12:25pm

No, the value of i is evaluated after all references to it have been "incremented". At the start of the your second call in the PRODUCT macro, NOT A FUNCTION CALL!, your code is actually k = (++i) * (++i). Both of the i variable references will be incremented BEFORE the expression starts to be evaluated. So your code is incrementing i twice, once for each mention of it, before the math expression is evaluated. Your assignment to k is actually k = 7 * 7, NOT k = 6 * 7.

Dave Kreskowiak 4-Nov-15 12:26pm

Here's the ASM code for it, if you care to read it:

_k$ = -32						; size = 4
_j$ = -20						; size = 4
_i$ = -8						; size = 4
_main	PROC						; COMDAT

; 9    : {

	push	ebp
	mov	ebp, esp
	sub	esp, 228				; 000000e4H
	push	ebx
	push	esi
	push	edi
	lea	edi, DWORD PTR [ebp-228]
	mov	ecx, 57					; 00000039H
	mov	eax, -858993460				; ccccccccH
	rep stosd

; 10   :     int i=3,j,k;

	mov	DWORD PTR _i$[ebp], 3			; Store 3 in i.

; 11   :     j=PRODUCT(i++);

	mov	eax, DWORD PTR _i$[ebp]     		; Copy i to EAX
	imul	eax, DWORD PTR _i$[ebp]			; Integer Multiply i by EAX (3 * 3) and store result in EAX.
	mov	DWORD PTR _j$[ebp], eax			; Copy EAX to J (value is 9).
	mov	ecx, DWORD PTR _i$[ebp]			; Copy i to ECX (value is 3).
	add	ecx, 1					; Add 1 to ECX.
	mov	DWORD PTR _i$[ebp], ecx			; Copy ECX to i (value is 4).
	mov	edx, DWORD PTR _i$[ebp]			; Copy i to EDX (value is 4).
	add	edx, 1					; Add 1 to EDX.
	mov	DWORD PTR _i$[ebp], edx			; Copy EDX to i (value is 5).

; 12   :     k=PRODUCT(++i);

	mov	eax, DWORD PTR _i$[ebp]			; Copy i to EAX (value is 5).
	add	eax, 1					; Add 1 to EAX.
	mov	DWORD PTR _i$[ebp], eax			; Copy EAX to i (value is 6).
	mov	ecx, DWORD PTR _i$[ebp]			; Copy i to ECX (value is 6).
	add	ecx, 1					; Add 1 to ECX.
	mov	DWORD PTR _i$[ebp], ecx			; Copy ECX to i (value is 7).
	mov	edx, DWORD PTR _i$[ebp]			; Copy i to EDX (value is 7).
	imul	edx, DWORD PTR _i$[ebp]			; Integer Multiply i by EDX (7 * 7) and store result in EDX.
	mov	DWORD PTR _k$[ebp], edx			; Copy EDX to k (value is 49).

; 13   :     printf("%d %d\r\n",j,k);

	mov	esi, esp
	mov	eax, DWORD PTR _k$[ebp]
	push	eax
	mov	ecx, DWORD PTR _j$[ebp]
	push	ecx
	push	OFFSET ??_C@_07BMEMJKJN@?$CFd?5?$CFd?$AN?6?$AA@
	call	DWORD PTR __imp__printf
	add	esp, 12					; 0000000cH
	cmp	esi, esp
	call	__RTC_CheckEsp