为什么我的 C 代码发生了分段错误:11

Question

我试图编写一个函数，该函数删除字符串s1中的每个字符，该字符与字符串s2中的任何字符匹配。这是挤压方法的测试代码。

#include <stdio.h>
void squeeze(char s1[], char s2[]);
int main()
{
    char s1[20] = "HelloWorld", s2[20] = "ol"; 
    squeeze(s1, s2); 
    printf("%sn", s1); 
    return 0; 
}
void squeeze(char s1[], char s2[])
{
    int i, j, k; 
    k = 0; 
    for (i = 0; s1[i] != ''; ++i) {
        for (j = 0; s2[j] != ''; ++j) {
            if (s1[i] != s2[j])
                s1[k++] = s1[i];
        }
    }
    s1[k] = '';
}

当我运行此代码时，终端总是给出Segmentation fault: 11.谁能给我任何提示为什么会发生这种情况？

Answer 1

示例：

#include <stdio.h>
static int found(char *str, char c) { //  return 1 if c is found in str
  for (size_t i = 0; str[i] != ''; i++) {
    if (str[i] == c) {
      return 1;
    }
  }
  return 0;
}
static void squeeze(char *a, char *b) {
  size_t k = 0;
  for (size_t i = 0; a[i] != ''; i++) { // use size_t to iterate on a c-string
    if (found(b, a[i]) != 1) { 
      a[k++] = a[i]; // copy only if a[i] is not in b
    }
  }
  a[k] = '';
}
int main(void) {
  char a[] = "HelloWorld"; // you should let auto size
  char b[] = "ol";         // and separate declaration
  squeeze(a, b);
  printf("%sn", a);
}

Answer 2

你的代码问题s1[k++] = s1[i];

尝试使用新阵列

请参阅以下代码

#include<stdio.h>
char *squeeze(char s1[], char s2[]);
int main()
{
    char s1[20] = "HelloWorld", s2[20] = "olH";
    squeeze(s1, s2);
    return 0;
}

char *squeeze(char s1[], char s2[])
{
    int i, j, k;
    k = 0;
    char arr[100]; // new array
    int flag = 0;
    for (i = 0; s1[i] != ''; i++) {
        flag = 0;
        for (j = 0; s2[j] != ''; ++j) {
            if (s1[i] == s2[j])
            {
                flag = 1;
                break;
            }
        }
        if (flag == 0)
        {
            arr[k++] = s1[i];
        }
    }
    arr[k] = '';
    printf("%s",arr);
}