C语言菜鸟问题:如下#include "stdio.h"void main(){ char a='A'+'8'-'4';\x05printf("%d",a);}这样算出来等于69,但是改为 char a='A'+'8'计算出来的是121?按我的理解:‘8’和‘4’计算时应当是计算其ASCII码而不是
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C语言菜鸟问题:如下#include "stdio.h"void main(){ char a='A'+'8'-'4';\x05printf("%d",a);}这样算出来等于69,但是改为 char a='A'+'8'计算出来的是121?按我的理解:‘8’和‘4’计算时应当是计算其ASCII码而不是
C语言菜鸟问题:如下
#include "stdio.h"
void main()
{ char a='A'+'8'-'4';
\x05printf("%d",a);
}
这样算出来等于69,但是改为 char a='A'+'8'计算出来的是121?
按我的理解:‘8’和‘4’计算时应当是计算其ASCII码而不是8和4(8的ASCII码是56),但是怎么会出现69呢?
C语言菜鸟问题:如下#include "stdio.h"void main(){ char a='A'+'8'-'4';\x05printf("%d",a);}这样算出来等于69,但是改为 char a='A'+'8'计算出来的是121?按我的理解:‘8’和‘4’计算时应当是计算其ASCII码而不是
#include <stdio.h>
int main(void)
{
char a = 'A' + '8' -'4';
char b = 'A' + '8';
char c = 'A';
char d = '8';
char e = '4';
printf("a = %d\n", a);
printf("b = %d\n", b);
printf("c = %d\n", c);
printf("d = %d\n", d);
printf("e = %d\n", e);
return 0;
}
这样打印出来你就知道了
'A' 等于十进制数值的65
‘8’ 等于十进制数值的56
‘4’ 等于十进制数值的52
然后进行十进制加减法吧