call printf using va
void TestPrint(char* format, ...) { va_list argList; va_start(argList, format); printf(format, argList); va_end(argList); } int main() { TestPrint("Test print %s %dn", "string", 55); return 0; }
I neet to get:
Test print string 55
Actually, I get garbage output. What is wrong in this code?
改用vprintf()
。
Instead of printf, I recommend you try vprintf instead, which was created for this specific purpose:
#include <stdio.h>
#include <stdlib.h>
#include <stdarg.h>
void errmsg( const char* format, ... )
{
va_list arglist;
printf( "Error: " );
va_start( arglist, format );
vprintf( format, arglist );
va_end( arglist );
}
int main( void )
{
errmsg( "%s %d %s", "Failed", 100, "times" );
return EXIT_SUCCESS;
}
Source: http://www.qnx.com/developers/docs/6.5.0/index.jsp?topic=/com.qnx.doc.neutrino_lib_ref/v/vprintf.html
As others have pointed out already: In this case you should use vprintf
instead.
But if you really want to wrap printf
, or want to wrap a function that does not have a v...
version, you can do that in GCC using the non-standard __builtin_apply
feature:
int myfunction(char *fmt, ...)
{
void *arg = __builtin_apply_args();
void *ret = __builtin_apply((void*)printf, arg, 100);
__builtin_return(ret);
}
The last argument to __builtin_apply
is the max. total size of the arguments in bytes. Make sure that you use a value here that is large enough.
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