C语言 警告：多字符字符常量 [-Wmultichar]|
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warning: multi-character character constant [-Wmultichar]|
提问by kri
/* Beginning 2.0 */
#include<stdio.h>
main()
{
printf(" %d signifies the %c of %f",9,'rise',17.0);
return 0;
}
Hello people
大家好
When I compile this, the compiler giving the following warning:
当我编译它时，编译器给出以下警告：
warning: multi-character character constant [-Wmultichar]|
And the output only prints einstead of rise.
并且输出只打印e而不是rise.
Are multiple characters not allowed in C?
C 中是否不允许使用多个字符？
How can I print the whole word (rise)?
如何打印整个单词 ( rise)？
Please help me out.
请帮帮我。
回答by Jerry Coffin
Try: printf(" %d signifies the %s of %f",9,"rise",17.0);.
试试：printf(" %d signifies the %s of %f",9,"rise",17.0);。
C distinguishes between a character (which is onecharacter) and a character string(which can contain an arbitrary number of characters). You use single quotes ('') to signify a character literal, but double quotes to signify a character stringliteral.
C 区分一个字符（一个字符）和一个字符串（可以包含任意数量的字符）。使用单引号 ( '') 表示字符文字，但使用双引号表示字符串文字。
Likewise, you specify %cto convert a single character, but %sto convert a string.
同样，您指定%c转换单个字符，但%s转换字符串。
回答by herohuyongtao
Use %sand ""for a character string:
将%s和""用于字符串：
printf(" %d signifies the %s of %f",9,"rise",17.0);
^^ ^ ^
For 'rise', this is valid ISO 9899:1999 C. It compiles without warning under gcc with -Wall, and a “multi-character character constant”warning with -pedantic.
对于'rise'，这是有效的ISO 9899：1999 C。它在 gcc 下编译时没有警告-Wall，“multi-character character constant”使用-pedantic.
According to the standard (§6.4.4.4.10),
根据标准（§6.4.4.4.10），
The value of an integer character constant containing more than one character (e.g., 'ab'), [...] is implementation-defined.
The value of an integer character constant containing more than one character (e.g., 'ab'), [...] is implementation-defined.
