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Views: 2872 | Replies: 4
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Join Date: Jul 2004
Posts: 31
Reputation: 
Rep Power: 5
Solved Threads: 0
Light Poster
Consider the following programme
void main()
{
char str[]="\12345s\n";
printf("%d",sizeof(str));
}
Output is :6
can someone explain me why.And is sizeof contain null value while
gigving the size of any string.
void main()
{
char str[]="\12345s\n";
printf("%d",sizeof(str));
}
Output is :6
can someone explain me why.And is sizeof contain null value while
gigving the size of any string.
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Join Date: Apr 2004
Posts: 3,755
Reputation: 
Rep Power: 17
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char str[] = "\12345s\n";
char str[] = {0123,'4','5','s','\n','\0'}; •
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Join Date: Jun 2004
Location: Marin, CA, USA
Posts: 434
Reputation:
Rep Power: 5
Solved Threads: 10
Unprevaricator
And if you said:
void main()
{
char *str ="\12345s\n";
printf("%d",sizeof(str));
}
The output will be 4, the size of a pointer (not the size of what it points to)
void main()
{
char *str ="\12345s\n";
printf("%d",sizeof(str));
}
The output will be 4, the size of a pointer (not the size of what it points to)
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Join Date: Jul 2004
Posts: 31
Reputation:
Rep Power: 5
Solved Threads: 0
Light Poster
char str[] = {0123,'4','5','s','\n','\0'};
how it is treated like this first element like 0123.it is very much confusing pls clear my doubt.
how it is treated like this first element like 0123.it is very much confusing pls clear my doubt.
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