snprintf using format specifier in the string getting copied

Question

I created the below program:

#include<stdio.h>
#include<stdlib.h>

#define TESTER "name=%s "

int main(){
    char *x;
    x = malloc(100);
    snprintf(x, 100, "Heyaa tester %s", TESTER, "hello");
    printf("%s", x);
    free(x)
    return 0;
}

I am basically trying to get the output something like - "Hey tester name=hello", However, it seems to be like below:

Heyaa tester name=%s

Do I need to append hello initially to the macro and then do snprintf to the malloc'd variable.

Thanks.

This will not work as you intended. This is practically `snprintf(x, 100, "Heyaa tester %s", "name=%s ", "hello");`. So you substitute the first `%s` with `"name=%s "` and the `"hello"` is basically ignored. What you could do is `snprintf(x, 100, "Heyaa tester"TESTER, "hello");` — Eraklon, Mar 27 '20 at 11:32

Inian · Answer 1 · 2020-03-27T11:44:30.007

4

You need to include the macro as part of the format specifier of snprintf() where it expands to "name=%s " at the time of pre-processor itself and the concatenated string "Heyaa tester name=%s" is created. What you have in the OP, makes name=%s as a literal string that will undergo %s formatting treatment by snprintf()

snprintf(x, 100, "Heyaa tester " TESTER, "hello");

Your original attempt should have been caught as suspicious, if you enabled extra warnings flag in your compiler as printf() would have thrown a warning too many arguments but not enough specifiers found ([-Wformat-extra-args] with gcc).

And as a better coding practice always clean up the memory allocated dynamically by the program instead of letting the OS doing it and always compile your program with extra warning flags.

edited Mar 27 '20 at 11:44

answered Mar 27 '20 at 11:35

Inian

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it is not the sollution as `char * TESTER = "name=%s ";` will not work. – 0___________ Mar 27 '20 at 11:48
1

@P__J__: Am sorry could you add more details to your comment. I can't see to understand what you are implying – Inian Mar 27 '20 at 11:50
@P__J__ Yes. I agree. It won't work with a variable. But here Macro is used. So it'd be fine. – RatDon Mar 27 '20 at 12:04

RatDon · Accepted Answer · 2020-03-29T05:25:02.500

2

As @Eraklon pointed out, though the Macro is expanded in the pre-processor, it's attached to the format specifier during compilation.

You can modify that line to the below and this will work.

snprintf(x, 100, "Heyaa tester " TESTER, "hello");

Here, the macro is expanded in pre-processor, then we are doing a string concatenation and after that the format specifier will pick up "hello" during compilation.

Just FYI: Macro and string, string and string, macro and macro can be concatenated just using a space in between. The compiler treats adjacent strings as single string. But beware it won't work with a string variable.

For further reading: C/C++ macro string concatenation

edited Mar 29 '20 at 05:25

answered Mar 27 '20 at 11:35

RatDon

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"Token pasting" is the use of the `##` preprocessor operator to create a new token. The rest of those things are not token pasting. And string concatenation is performed *after* preprocessing. – rici Mar 27 '20 at 13:43
@rici Thanks for the info. Edited the answer. – RatDon Mar 29 '20 at 05:23

snprintf using format specifier in the string getting copied

2 Answers2