通常函数的用法:
我能使用这个参数吗?
my_func("test");
我能使用这个参数吗?
my_func(printf("current_count_%d",ad_id));
int my_func(const char *key)
my_func("test");
my_func(printf("current_count_%d",ad_id));
int my_func(const char *key)
是的,您可以将printf
的返回值用作函数参数。
但请记住,printf
成功时返回写入的字符数。
因此
foo(printf("bar%d",123));
函数将6
作为参数传递给foo
,而不是bar123
。
如果您想传递printf
打印的字符串,可以使用sprintf
函数。它类似于printf
,但不是写入控制台,而是写入到字符数组中。
char buf[64]; /* malloc or whatever */
int pos = snprintf(buf, sizeof(buf), "current_count_%d", ad_id);
if (sizeof(buf) <= pos)
buf[sizeof(buf)-1] = '\0';
my_func(buf)
void varargfunc(char *fmt, ...)
{
char formattedString[2048]; /* fixed length, for a simple example - and
snprintf will keep us safe anyway */
va_list arguments;
va_start(arguments, fmt); /* use the parameter before the ellipsis as
a seed */
/* assuming the first argument after fmt is known to be an integer,
you could... */
/*
int firstArgument = va_arg(arguments, int);
fprintf(stderr, "first argument was %d", firstArgument);
*/
/* do an vsnprintf to get the formatted string with the variable
argument list. The v[printf/sprintf/snprintf/etc] functions
differ from [printf/sprintf/snprintf/etc] by taking a va_list
rather than ... - a va_list can't turn back into a ... because
it doesn't inherently know how many additional arguments it
represents (amongst other reasons) */
vsnprintf(formattedString, 2048, fmt, arguments);
/* formattedString now contains the first 2048 characters of the
output string, with correct field formatting and a terminating
NULL, even if the real string would be more than 2047 characters
long. So put that on stdout. puts adds an additional terminating
newline character, so this varies a little from printf in that
regard. */
puts(formattedString);
va_end(arguments); /* clean up */
}
不,printf()
返回打印到标准输出的字符数。使用s[n]printf()
创建字符串,然后传递该字符串。
char[] buffer
并在使用完毕后释放它。 - Christian Mannprintf
函数返回一个整数。
int printf ( const char * format, ... );
my_func
接受整数作为参数,就可以在my_func
中使用它。但事实似乎并非如此。
my_func
的意图是什么? - Keynslug