Question

Possible Duplicate:
Why does passing the result of printf to another printf work?

I have a code snippet..

printf("%d", printf("tim"));

The function printf prints the value, tim3 .. The second printf statement does not have a specifier so why do the number of characters get printed along with the string “tim” ?

When i only run this code .. printf("tim"); i get output as, tim Exited: ExitFailure 3 Why does this happen?

And how does the 1st printf statement takes printf("tim") as an argument when it is expecting an integer?

Answer 1

printf prints the string as it is if no format specifier is specified. So printf("Hello"); will print Hello as it is.

You can also do the same by using a format specifier like so – printf("%s", "Hello");

printf also returns the number of characters printed. So printf("Hello"); first prints the string Hello and then returns 5.

In your statement, you’re printing the return statement of printf using printf("%d", ...);

In effect, the statement that you’ve given can be written like this –

int i = printf("tim");
printf("%d", i);

As for the failure, I’m guessing you have a return printf("tim"); in your main function.