Some C compilers permit multiple characters in a character constant. This means that writing 'yes' instead of "yes" may well go undetected. Source: C traps and pitfalls
任何人都可以举一个字符常量中允许多个字符的例子吗?
最佳答案
正如 Code Monkey 所引用的那样,它是由实现定义的并且实现各不相同——这不仅仅是 BigEndian/LittleEndian 和字符集的差异。我已经用程序测试了四种实现(全部使用 ASCII)
#include <stdio.h>
int main()
{
unsigned value = 'ABCD';
char* ptr = (char*)&value;
printf("'ABCD' = %02x%02x%02x%02x = %08x\n", ptr[0], ptr[1], ptr[2], ptr[3], value);
value = 'ABC';
printf("'ABC' = %02x%02x%02x%02x = %08x\n", ptr[0], ptr[1], ptr[2], ptr[3], value);
return 0;
}
我得到了四个不同的结果
大端(AIX、POWER、IBM 编译器)
'ABCD' = 41424344 = 41424344
'ABC' = 00414243 = 00414243
大端(Solaris、Sparc、SUN 编译器)
'ABCD' = 44434241 = 44434241
'ABC' = 00434241 = 00434241
小端(Linux、x86_64、gcc)
'ABCD' = 44434241 = 41424344
'ABC' = 43424100 = 00414243
小端(Solaris、x86_64、Sun 编译器)
'ABCD' = 41424344 = 44434241
'ABC' = 41424300 = 00434241
关于c - 字符常量中的多个字符,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59439824/