Question

我想要的只是用户输入的foo数组，并从其左右打印操作。

我尝试添加以下代码，但无法获得相同的结果。

我不是写char foo [29] = "1021+2551";而是这样写：

int i;

printf("Enter the number and opperator");

for(i=0; i<1; i++)      // I used i < 1 b/c I want to get only one line input

{

scanf("%c", foo[i]);

}

void main(){

    int index;
    char foo[29] = "1021+2551";
    int len = strlen(foo);

    for (int i=0; i < len; i++)
    {
    if(foo[i] == '+' || foo[i] == '-' || foo[i] == '*' || foo[i] == '/' || foo[i] == '%'){
       char op = foo[i];
       printf("%c", op);
       index = i;
        }
    }
    char left;
    for(int j=0; j < index; j++){
       left = printf("%c",foo[j]);
    }
    char right;
    for(int k=index + 1; k < len; k++){
       right = printf("%c",foo[k]);
    }
}

如果我在用户输入中输入100 + 200，则结果应为

+100200

Answer 1

简单而又快速的事情：）

    int main(){

    int i;
    char foo[29];

    printf("Enter the number and opperator\n");
    fgets(foo, 29 , stdin);

    int index;
    int len = strlen(foo);

    for (int i=0; i < len; i++)
    {
    if(foo[i] == '+' || foo[i] == '-' || foo[i] == '*' || foo[i] == '/' || foo[i] == '%'){
       char op = foo[i];
       printf("%c", op);
       index = i;
        }
    } printf(" ");
    char left;
    for(int j=0; j < index; j++){
       left = printf("%c",foo[j]);
    } printf(" ");
    char right;
    for(int k=index + 1; k < len; k++){
       right = printf("%c",foo[k]);
    }
    return 0;
}

如何使数组从用户读取？

1 个答案: