a[i++] = j

[saurabh29789]

Is the behavior of

a[i++] = j;

and/or

a[++i] = j;

undefined??

 

[Kojak]

Le Fri, 20 Feb 2009 09:33:03 -0800 (PST),
saurabh29789 a écrit :

Is the behavior of

a[i++] = j;

and/or

a[++i] = j;

undefined??

Why should it be undefined ? It's perfectly legal.

Juste keep in mind that 2 lines differs slightly, depending
on what you expect (i++ vs. ++i).

 

[jameskuyper]

Is the behavior of

a[i++] = j;

and/or

a[++i] = j;

undefined??

That depends very much upon what 'a', 'i', and 'j' are. There's no
reason within either of those statements themselves why the behavior
should be undefined. However, consider:

#include <limits.h>

char *i = "Hello!"+6;
int a = 1;
float j = CHAR_MAX*2.0F;

With those definitions "a[i++] = j;" has undefined behavior for at
least two different reasons. If a, i, or j are macros, the list of
different ways in which something could go wrong is practically
endless.

 

[Kenny McCormack]

a[i++] = j;
a[++i] = j;
undefined??

More context is needed. Imagine sticking the following before
the above:

#define j i

Or even: struct { int a; } i;

 

[Fred]

a[i++] = j;
a[++i] = j;
undefined??

More context is needed. Imagine sticking the following before
the above:

   #define j i

Or even: struct { int a; } i;

Or many other things such as
void a();
double i[10];