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发表于 2017-5-18 18:22:06 | 显示全部楼层 |阅读模式

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请问一下在ubuntu14.04上出现了一下报错信息是什么情况?
代码和报错信息如下:
#include <stdio.h>

int main()
{
        int i;
        char j;
        float k;

        i = 123;
        j = 'C';
        k = 3.14;

        printf("size of int is %d\n", sizeof(int));
        printf("size of i is %d\n", sizeof(i));
        printf("size of char is %d\n", sizeof(char));
        printf("size of j is %d\n", sizeof j);
        printf("size of float is %d\n", sizeof(float));
        printf("size of k is %d\n", sizeof k);

        return 0;
}



在LINUX下运行报错信息如下:
root@tc:~/C/s1e6# gcc work1.c -o work1
work1.c: In function ‘main’:
work1.c:12:2: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
  printf("size of int is %d\n",sizeof(int));
  ^
work1.c:13:2: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
  printf("size of i is %d\n",sizeof(i));
  ^
work1.c:14:2: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
  printf("size of char is %d\n",sizeof(char));
  ^
work1.c:15:2: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
  printf("size of j is %d\n",sizeof(j));
  ^
work1.c:16:2: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
  printf("size of float is %d\n",sizeof(float));
  ^
work1.c:17:2: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
  printf("size of k is %d\n",sizeof k);
  ^
最佳答案
2017-5-18 18:28:22
本帖最后由 超凡天赐 于 2017-5-18 18:29 编辑

提示就是让你把类型改为无符号长整型。
我认为你这个只是警告。我猜和国际标准有关。你可以查一下c99和c11标准。下面代码应该不会警告。
#include <stdio.h>
int main()
{
    int i;
    char j;
    float k;
    i = 123;
    j = 'C';
    k = 3.14;
    printf("size of int is %lu\n", sizeof(int));
    printf("size of i is %lu\n", sizeof(i));
    printf("size of char is %lu\n", sizeof(char));
    printf("size of j is %lu\n", sizeof j);
    printf("size of float is %lu\n", sizeof(float));
    printf("size of k is %lu\n", sizeof k);
    return 0;
}
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发表于 2017-5-18 18:28:22 | 显示全部楼层    本楼为最佳答案   
本帖最后由 超凡天赐 于 2017-5-18 18:29 编辑

提示就是让你把类型改为无符号长整型。
我认为你这个只是警告。我猜和国际标准有关。你可以查一下c99和c11标准。下面代码应该不会警告。
#include <stdio.h>
int main()
{
    int i;
    char j;
    float k;
    i = 123;
    j = 'C';
    k = 3.14;
    printf("size of int is %lu\n", sizeof(int));
    printf("size of i is %lu\n", sizeof(i));
    printf("size of char is %lu\n", sizeof(char));
    printf("size of j is %lu\n", sizeof j);
    printf("size of float is %lu\n", sizeof(float));
    printf("size of k is %lu\n", sizeof k);
    return 0;
}

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