printf似乎忽略了字符串精度-Java 学习之路

所以，我有点受阻 . 根据我系统上的 man 3 printf ，字符串格式 "%5s" 应该使用指定的精度来限制从给定的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.

但是我的测试代码没有证实这一点：

#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

当我想我应该得到

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

我做错了什么，还是手册对我说谎？

仅供参考：我知道我可以（通常）破解字符串，并插入临时 '\0' 来终止字符串（除非它是 char const * ，就像这里一样，我是'd have to copy it instead), but it'是PITA（特别是如果我试图打印两半的东西）相同的printf），我想知道为什么忽略精度 .

1 回答

19
您没有设置精度，而是设置字段宽度 . 精度始终以格式规范中的 . 开头 .
```
printf("test: %.3s\n", test);
```
回复于 2024-04-27T14:13:13+08:00

printf似乎忽略了字符串精度

1 回答

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