Is there a function around somewhere that I can use to predict the space that sprintf( ) will need? IOW, can I call a function size_开发者_StackOverflowt predict_space( "%s\n", some_string ) that will return the length of the C-string that will result from sprintf( "%s\n", some_string )?
In C99 snprintf
(note: Windows and SUSv2, do not provide an implementation of snprintf (or _snprintf) conforming to the Standard):
7.19.6.5 The snprintf function Synopsis [#1] #include <stdio.h> int snprintf(char * restrict s, size_t n, const char * restrict format, ...); Description [#2] The snprintf function is equivalent to fprintf, except that the output is written into an array (specified by argument s) rather than to a stream. If n is zero, nothing is written, and s may be a null pointer. Otherwise, output characters beyond the n-1st are discarded rather than being written to the array, and a null character is written at the end of the characters actually written into the array. If copying takes place between objects that overlap, the behavior is undefined. Returns [#3] The snprintf function returns the number of characters that would have been written had n been sufficiently large, not counting the terminating null character, or a negative value if an encoding error occurred. Thus, the null- terminated output has been completely written if and only if the returned value is nonnegative and less than n.
For example:
len = snprintf(NULL, 0, "%s\n", some_string);
if (len > 0) {
newstring = malloc(len + 1);
if (newstring) {
snprintf(newstring, len + 1, "%s\n", some_string);
}
}
Use can use snprintf() with a size of of 0 to find out exactly how many bytes will be required. The price is that the string is in effect formatted twice.
You can use snprintf
for that, as in
sz = snprintf (NULL, 0, fmt, arg0, arg1, ...);
But see Autoconf's portability notes on snprintf
.
In most cases, you can compute it by adding length of the string you are concatenating and taking max length for numeric values based on the format you used.
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