On 3/26/06, Shriramana Sharma <samjnaa@xxxxxxxxx> wrote:
> Sunday, 26 March 2006 16:27 samaye, Steve Graegert alekhiit:
>
> > Because the format (the string to display) is provided as a constant
> > expression and it is not being modified:
> > printf("%s\n", "abc");
> > printf("%s\n", mystring);
> > In both cases the format argument is a constant string.
>
> Why, the following works as well:
>
> #include "stdio.h"
> void main(void)
> {
> char s[10] = "\n%s\n\n";
> printf(s, "hello");
> }
>
> Here s is not a const char *. It is a variable char *. That actually
> compiled and executed.
Yes, it compiles and why should it not do so? It's a __string constant__, since it cannot be changed in any way. It's effectively the same as the statement
char *s = "\n%s\n\n"
The situation is different with malloc(2). It allows for dynamic allocation of memory, thus turning string constants into dynamic data structures:
char * s;
s = (char *) malloc(100);
Every string inside double quotes (" ") is a character constant.
\Steve
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