fwprintf: only the first character from a wide char array argument gets copied to output

Question

I'm trying to write a message containing "alpha = abcd" to a text file using the following code:

#include <stdio.h>
#include <wchar.h>

int
main()
{
        const wchar_t *a = L"abcd", *msg = L"alpha = %s\n";
        FILE          *f = fopen("./deleteme", "a");

        fwprintf(f, msg, a);
        fclose(f);
}

However, after compiling and executing the program I get this output instead:

alpha = a

Why only the first character from the const a gets copied to output?

score 2 · Accepted Answer · answered Sep 25 '17 at 12:38

You need to change:

L"alpha = %s\n";

to:

L"alpha = %S\n";

The argument (a = L"abcd") you are trying to print is a wide string, and therefore you need %S (uppercase) instead of %s (lowercase). Using incorrect format specifiers for printf-like functions is undefined behaviour.

Read the documentation for printf format specifiers.

score 1 · Answer 2 · answered Sep 25 '17 at 14:17

1

With a C99 compliant compiler , use "%ls".

If an l length modifier is present, the argument shall be a pointer to the initial element of an array of wchar_t type. C11dr §7.29.2.1 10.

// const wchar_t *msg = L"alpha = %s\n";
const wchar_t *msg = L"alpha = %ls\n";

answered Sep 25 '17 at 14:17

chux - Reinstate Monica

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@Kais Does this work for your compiler? If not, what compiler/version are you using? – chux - Reinstate Monica Sep 25 '17 at 14:37
Yes. BTW, I'm using gcc 6.3.0 under linux. – user6039980 Sep 25 '17 at 21:35
@Kais I use `6.4.0`, yet the issue may be more of the library version: [Ref](https://unix.stackexchange.com/questions/120380/what-c-library-version-does-my-system-use) - mine is 2.9.0 – chux - Reinstate Monica Sep 25 '17 at 21:43
It might be the case .. mine is 2.24. – user6039980 Sep 25 '17 at 21:49

fwprintf: only the first character from a wide char array argument gets copied to output

2 Answers2