Question

基本上我试图从用户那里得到一个字符串：3 + 2，作为单个字符串，单独检查字符串的组成部分并进行计算。我已经让计算器程序使用命令行输入，但如果他们不使用命令行，我需要从用户获取和读取字符串输入

这是我到目前为止所拥有的。请hellppp :(

#include <stdio.h>
#include <math.h>



int main(int argc, char *argv[]) {
int num1, num2;
float ans;
char operator;
int i = 0, j = 0;
char s[100] = " ";
char s1[100], s2[100];




if (argc==4) {
sscanf(argv[1], "%d", &num1);
sscanf(argv[2], "%c", &operator);
sscanf(argv[3], "%d", &num2);

switch (operator) {
  case '+': ans = num1+num2;
    printf("%d %c %d = %.2f\n", num1, operator, num2, ans);
        break;
  case '-': ans = num1-num2;
    printf("%d %c %d = %.2f\n", num1, operator, num2, ans);
        break;
  case 'x':
  case 'X': ans = num1*num2;
        printf("%d %c %d = %.2f\n", num1, operator, num2, ans);
            break;
  case '^': ans = pow(num1, num2);
            printf("%d %c %d = %.2f\n", num1, operator, num2, ans);
            break;

  case '/': if (num2 == 0) {
      printf("Error! Division by Zero!\n");
    }
    else {
      ans = num1/num2;
      printf("%d %c %d = %.2f\n", num1, operator, num2, ans);
    }           
        break;
  default: printf("%c is not a valid operator!\n", operator);
  }}
 else
 if (argc==3) {
sscanf(argv[1], "%c", &operator);
sscanf(argv[2], "%d", &num1);

switch (operator) {
  case 'n':
  case 'N': ans = -num1;
        printf("-(%d) = %.2f\n", num1, ans);
        break;
  case 'a':
  case 'A': ans = fabs(num1);
        printf("|%d| = %.2f\n", num1, ans);
    break;


  case 's':
  case 'S': if (num1 < 0) {
      printf("Can't find square root of negative number\n");
    }
    else {
      ans = sqrt(num1);
      printf("Sqrt(%d) = %.2f\n", num1, ans);
    }
        break;
  default: printf("%c is not a valid operator!\n", operator);
} }

else if ((argc!=3) || (argc!=4)) {

printf("Enter the string you want to calculate: ");
gets(s);




while (s[i] != '\0') {

s[i] = argv[1];

i++;

} 

 while 
(s[i] != '\0') {

s[i] = argv[2];

i++;

}
 while 
 (s[i] != '\0') {

 s[i] = argv[3];

 i++;

 }}

}
else 
printf("Input Error!!!\n");
}

Answer 1

首先，您应该使用fgets来读取输入，因为gets容易受到缓冲区溢出的影响：

fgets(s, sizeof(s), stdin);

然后，您只需使用sscanf即可获得所需的数据：

sscanf(s, "%d %c %d", &num1, &operator, &num2);

在C中读取计算器的字符串输入

1 个答案: