从'char'无效转换为'char *'[-fpermissive] strcpy（choice，argv [1]）;

Question

对有理数的操作，我要从stdin和命令行获得输入。当我尝试将argv值cpy到变量时，我得到错误：从'char'无效转换为'char *'[-fpermissive] strcpy（choice，argv [1]）;

    #include <stdio.h>
    #include<string.h>
    typedef struct rational
    { int num,den;} Rational; 
    Rational rationalNumMain(int argc, const char** argv)
    {
        Rational r1,r2,r;char choice;
        if(argc==1)
        {
        scanf("%c",&choice);
        scanf("%d",&r1.num);
        scanf("%d",&r1.den);
        scanf("%d",&r2.num);
        scanf("%d",&r2.den);
        }
        else
        {  strcpy(choice,argv[1]);
        strcpy(r1.num,argv[2]);
        strcpy(r1.den,argv[3]);
        strcpy(r2.num,argv[4]);
        strcpy(r2.den,argv[5]);
        }
    switch(choice)
    {
    case 'A':
    r.num=(r1.num*r2.den)+(r1.den*r2.num);
    r.den=(r1.den*r2.den);
    return r;
    break;
    case 'S':
    r.num=(r1.num*r2.den)-(r1.den*r2.num);
    r.den=(r1.den*r2.den);
    return r;
    break;
    case'M':
    r.num=(r1.num*r2.num);
    r.den=(r1.den*r2.den);
    return r;
    break;
    case 'D':
    r.num=(r1.num*r2.den);
    r.den=(r1.den*r2.num);
    return r;
    break;
   } }
    int main(int argc, const char** argv)
    {
    Rational r1,r2,result;
    result=rationalNumMain(argc, argv);
    printf("%d,%d", result.num, result.den);//prints the output
    return 0;
    }

Answer 1

使用sscanf解析来自argv的输入。

sscanf(argv[1], "%c", &choice);
sscanf(argv[2], "%d",&r1.num);
sscanf(argv[3], "%d",&r1.den);
sscanf(argv[4], "%d",&r2.num);
sscanf(argv[5], "%d",&r2.den);

Answer 2

'choice'被声明为char而不是char *所以这将失败

strcpy(choice,argv[1]);

因为strcpy将两个字符串指针作为参数。

为了从argv取消引用它的第一个字符

choice = *argv[1];