C: 分配给 char... 期待类型转换？

Question

使用

int main(int argc, char *argv[])

{
    double number1, number2;
    char operator;

    number1 =atof (argv[0]);
    operator =argv[1];        // line 29
    number2 =atof (argv[2]);

编译器提示说

29 warning: assignment makes integer from pointer without a cast [enabled by default]

Answer 1

argv[1] 是指向 char 的指针，您不能将其分配给 char。将运算符转换为 char * 或尝试 operator = *(argv[1]);