printf 似乎忽略了字符串精度

Question

所以，我有点难受。根据我系统上的 man 3 printf，字符串格式 "%5s" 应使用指定的精度来限制从给定字符串参数打印的字符数。

% man 3 printf
PRINTF(3)                BSD Library Functions Manual                PRINTF(3)

NAME
     printf, fprintf, sprintf, snprintf, asprintf, vprintf, vfprintf,
     vsprintf, vsnprintf, vasprintf -- formatted output conversion

...
     s       The char * argument is expected to be a pointer to an array of
             character type (pointer to a string).  Characters from the array
             are written up to (but not including) a terminating NUL charac-
             ter; if a precision is specified, no more than the number             
             specified are written.  If a precision is given, no null
             character need be present; if the precision is not specified, or
             is greater than the size of the array, the array must contain a
             terminating NUL character.

但我的测试代码并没有证实这一点：

#include <stdio.h>
int main()
{
        char const * test = "one two three four";
        printf("test: %3s\n", test);
        printf("test: %3s\n", test+4);
        printf("test: %5s\n", test+8);
        printf("test: %4s\n", test+14);
        return 0;
}

它输出

test: one two three four
test: two three four
test: three four
test: four

When I think I should be getting

test: one
test: two
test: three
test: four

Am IDothing bad, 或者手册页只是在骗我？

仅供参考：我知道我（通常）可以破解字符串，并插入临时 '\0' 来终止字符串（除非它是 char const * ，如下所示，我必须复制它），但它是一个 PITA（特别是如果我试图在同一个 printf 中打印某些东西的两半），并且我想知道为什么精度被忽略。

Answer 1

您没有设置精度，而是设置字段宽度。在格式规范中，精度始终以. 开头。

printf("test: %.3s\n", test);