Cannot able to understand the output

Code:

---

#include<stdio.h>
void main()
{
 printf("%%%%\n");
}

---

output:
%%

Why the output is '%%'??
Why not '%%%%'?
If I replace '%' with '$', then output is '$$$$'. Why?

I am using Turbo C 4.5 on Windows 7 platform

'%' is a format specifier. It has special meaning. Just like '\' is expecting more characters afterwards but if you want '\' output then you have to use '\\'.

Quote:

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Originally Posted by raj.knd

I am using Turbo C 4.5 on Windows 7 platform

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My condolences...

(You really should think about a better compiler, something like ... Pelles C)

Yes. I made the changes and got the answer. Thank you.

Thank you Tater. I will try to get that one.

you should also know that the return type of main is int not void.and you can't get away with void main() in up-to-date compilers like pelles c. read this
Cprogramming.com FAQ > main() / void main() / int main() / int main(void) / int main(int argc, char *argv[])

the printf() function uses the % symbol to identify the conversion specifications, there is a slight problem if you want to print the % sign itself. If you simply use a lone % sign, the compiler thinks you have bungled a conversion specification. The way out is simple—just use two % symbols, as shown here:


pc = 2*6; printf("Only %d%% of Sally's gribbles were edible.\n", pc);
The following output would result:



Only 12% of Sally's gribbles were edible.