#include <stdio.h> int main() { char c = 'a'; int n = c ; //将c赋值给n float f = c; //将c赋值给f double d = c; //将c赋值给d printf("%d\n",n); printf("%f\n",f); printf("%lf\n",d); return 0; }

源自：2-8 自动类型转换

#include <stdio.h> int main() { char c = 'a'; int n = c ; //将c赋值给n float f = c; //将c赋值给f double d = c; //将c赋值给d printf("%d\n",n); printf("%f\n",f); printf("%lf\n",d); return 0; }

第二个输出不是单精度吗？为啥运行结果是97.000000

提问者：慕莱坞4462739 2017-08-03 17:57

个回答

qq_陌_34: 2017-08-03 18:45:21

这里的精度指的是有效数字的位数，后面的零都是为了区别于整型的，所以默认是那么多个，

ICQ小乐乐

2017-08-03 18:37:16

#include <stdio.h>
int main()
{
    char c = 'a';
    int n = c ; //将c赋值给n
    float f = c; //将c赋值给f
    double d = c; //将c赋值给d
    printf("%d\n",n); //97
    printf("%f\n",f); //97.000000
    printf("%lf\n",d);//97.000000000000000
    return 0;
}
单精度(float--%f)精确到小数点后第6位，双精度(double--lf)精确到小数点后15位

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qq_万人呐喊我耀宗_0

2017-08-03 18:37:05

#include <stdio.h>
int main()
{
char c = 'a';
int a =97;
int n = c; //将c赋值给n
float f = c; //将c赋值给f
double d = c; //将c赋值给d
printf("%d\n",n);
printf("%f\n",f);
printf("%lf\n",d);
return 0;
}
%f 在前面讲解过小数的

0