当输出足够小时，给Snprintf（）尺寸太大
此代码会导致不确定的行为吗？因为缓冲区只有128个字节，但我告诉snprintf（）它的时间更长。但是，所得的字符串比128个字节短。
#include <stdio.h>
int main(void)
{
char buffer[128];
snprintf(buffer,294201,"%s","ABC");
puts(buffer);
return 0;
}
Does this code cause undefined behaviour? Because the buffer is only 128 byte long but i tell snprintf() that it is longer. However, the resulting string is shorter than 128 byte.
#include <stdio.h>
int main(void)
{
char buffer[128];
snprintf(buffer,294201,"%s","ABC");
puts(buffer);
return 0;
}
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C 2018 7.21.6.5 2说：
请注意，这不说
snprintf传递了n或更多字符的数组。因此，snprintf没有任何许可证可以假定它可以写入s [n-1]，除非fprintf等同于写作n字符（包括终止空字符）。换句话说，假设我们定义一个数组
缓冲区294,201个字符，用数据填充它，然后调用snprintf（buffer，294201，“％s”，“ abc”）; <<< /代码>。除了前四个字符之外，我们会期望什么都不会改变吗？如果缓冲区中的其他一些字节更改，则此snprintf调用将不等于fprintf，只是将输出写入数组中……”如果该规范在缓冲区中更改了任何内容，则违反了该规范。C 2018 7.21.6.5 2 says:
Note this does not say
snprintfis passed an array ofnor more characters. Sosnprintfis not given any license to assume it may write tos[n-1]unless thefprintfthat it is equivalent to would writencharacters (including the terminating null character).Looking at this another way, suppose we define an array
bufferof 294,201 characters, fill it with data, and callsnprintf(buffer,294201,"%s","ABC");. Would we expect nothing beyond the first four characters to change? If some other byte in the buffer changed, then thissnprintfcall would not be “equivalent tofprintf, except that the output is written into an array…” I would deem it a violation of this specification if it changed anything further in the buffer.