有没有办法使用printf()指定要打印的字符串的字符数？

Question

有没有办法指定要打印的字符串的字符数(类似于ints中的小数位)？

printf ("Here are the first 8 chars: %s\n", "A string that is more than 8 chars");

想要打印: Here are the first 8 chars: A string

Answer 1

基本方法是:

printf ("Here are the first 8 chars: %.8s\n", "A string that is more than 8 chars");

另一种通常更有用的方法是:

printf ("Here are the first %d chars: %.*s\n", 8, 8, "A string that is more than 8 chars");

在这里,您将length指定为printf()的int参数,它将格式中的'*'视为从参数获取长度的请求.

您也可以使用表示法:

printf ("Here are the first 8 chars: %*.*s\n",
        8, 8, "A string that is more than 8 chars");

这也类似于"%8.8s"符号,但同样允许您在运行时指定最小和最大长度 - 在以下情况中更实际:

printf("Data: %*.*s Other info: %d\n", minlen, maxlen, string, info);

POSIX规范printf()定义了这些机制.

Answer 2

printf ("Here are the first 8 chars: %.8s\n", "A string that is more than 8 chars");

%8s将指定最小宽度为8个字符.您希望截断为8,因此请使用%.8s.

如果您想要始终打印8个字符,则可以使用%8.8s

Answer 3

用printf你可以做到

printf("Here are the first 8 chars: %.8s\n", "A string that is more than 8 chars");

如果您使用的是C++,则可以使用STL获得相同的结果:

using namespace std; // for clarity
string s("A string that is more than 8 chars");
cout << "Here are the first 8 chars: ";
copy(s.begin(), s.begin() + 8, ostream_iterator<char>(cout));
cout << endl;

或者,效率较低:

cout << "Here are the first 8 chars: " <<
        string(s.begin(), s.begin() + 8) << endl;

Answer 4

除了指定固定数量的字符外,您还可以使用*这意味着printf从参数中获取字符数:

#include <stdio.h>

int main(int argc, char *argv[])
{
        const char hello[] = "Hello world";
        printf("message: '%.3s'\n", hello);
        printf("message: '%.*s'\n", 3, hello);
        printf("message: '%.*s'\n", 5, hello);
        return 0;
}

打印:

message: 'Hel'
message: 'Hel'
message: 'Hello'

Answer 5

printf(....."%.8s")