Float被视为双重

Question

C使用Xcode 运行这个小脚本时,我收到以下消息:

Format specifies type 'float *' but the argument has type 'double" at scanf("%f", v) and scanf("%f", i).

我没有得到它,因为我没有声明任何double类型变量.

int main(int argc, const char * argv[]) {
    char choice[10];

    float v;
    float i;
    float r;

    printf("What would you like to calculate?: ");
    scanf("%s", choice);
    printf("\nYou chose: \n""%s", choice);

    if (strcmp(choice, "r") == 0)
    { 
        printf("\nPlease enter voltage (V): \n");
        scanf("%f", v);

        printf("\nPlease enter current (I): \n");
        scanf("%f", i);

        r = v/i;

        printf("%f", r);
    }
}

有任何想法吗？

Answer 1

您收到该警告是因为您未能将指向float(float*)的指针传递给该函数scanf.编译器告诉你它是double,因为它scanf是一个可变参数函数.变量参数受默认参数提升的约束,其中某些数据类型的参数被转换为更大的数据类型.在这种情况下,float被提升为double.

在C函数的唯一方法修改变量v,i以及choice是通过他们为指针,所以你需要传递指针scanf,使用&运营商"的地址".

您的代码应如下所示:

int main(int argc, const char * argv[]) {
    char choice[10];

    float v;
    float i;
    float r;

    printf("What would you like to calculate?: ");
    scanf("%9s", &choice); /* this specifier prevents overruns */
    printf("\nYou chose: \n""%s", choice);

    if (strcmp(choice, "r") == 0)
    { 
        printf("\nPlease enter voltage (V): \n");
        scanf("%f", &v); /* use a pointer to the original memory */

        printf("\nPlease enter current (I): \n");
        scanf("%f", &i); /* use a pointer to the original memory */

        r = v/i;

        printf("%f", r);
    }
}

另请注意,我使用了格式说明符%9s.这样,如果用户输入的字符超过9个,则不会覆盖相邻的存储器.您必须为null字符保留数组的最后一个元素,\0因为C中的字符串以\0.