在C编程中将元素插入到动态字符数组中

问题描述

尝试在C编程中添加元素到动态字符数组时，我有一些问题。以下是预期输出：

 您要输入的字符数：5 
输入字符串：datas 
该字符串为：datas 
要插入或2删除或3-quit？：1 
要插入的字符是什么：a 
生成的字符串：adata 
  

我已经在主要功能中做了这些用户输入部分，这里是主要的代码我输入字符串输入，大小并传递给insert（）：

  printf（你想要多少个字符输入：）; 
 scanf（％d，& n）; 
 str = malloc（n + 1）; 
 printf（输入字符串类：）; 
 scanf（％s，str）; 
 
 case'1'：
 printf（你要插入的是什么字符）; 
 scanf（％c，& input）; 
 insert（str，input，n）; 
 break; 
  

而我的insert（）：

的部分

  void insert（char * str，char input，int n）{
 int i; 
 size_t space = 1; （i = 0; i  str [i] =（char）（input + i）; 
 
 space ++; 
 str = realloc（str，space）; 
 if（i> 2）{
 break; （i = 0; i< n; i ++）{
 printf（％c，str [i]）; 
} 
} 
 
 
} 
} 
  

但是，当我尝试打印出来的字符串insert（），假设我输入'a'附加到动态数组的第一个元素，大小为5，我得到的结果是<$ c $我从，我不知道如何解决这个问题。感谢提前。

解决方案

这是代码 - 与合同，呼叫者做空闲的一点！调用者使用 insert（& str，input，n）调用它

  void insert（char ** str，char input，int n）{
 
 char * temp = * str; 
 int i; 
 
 * str = realloc（* str，n + 2）; / * realloc first * / 
 
 if（！* str）/ * realloc failed * / 
 {
 fputs（realloc failed，stderr）; 
 free（temp）; / *释放以前的malloced内存* / 
 exit（-1）; / *退出程序* / 
} 
 
 for（i = n; i> = 0; i--）{
（* str）[i + 1] =（* str）[i]; / *将所有字符向上移动* / 
} 
 
（* str）[0] =输入; / *插入新的字符* / 
 
 printf（％s，* str）; / *打印新的字符串* / 
} 
  

对于格式化很抱歉。这是留给读者的。我没有检查算法，但这并不会泄漏记忆

I was having some problem when trying to add element to dynamic char array in C programming. Here is the expected output:

How many characters do you want to input: 5
Input the string:datas
The string is: datas
Do you want to 1-insert or 2-remove or 3-quit?: 1
What is the character you want to insert: a
Resulting string: adata

I already did those user input part in the main function and here is the code in main where I take in the string input, size and pass them to insert():

printf("How many characters do you want to input: ");
scanf("%d", &n);
str = malloc(n + 1);
printf("Input the string class: ");
scanf("%s", str);

case '1':
    printf("What is the character you want to insert: ");
    scanf(" %c", &input);
    insert(str, input, n);
    break;

And the part where my insert():

void insert(char *str, char input, int n) {
int i;
size_t space = 1;
for (i = 0; i < n; i++) {
    str[i] = (char)(input + i);
    space++;
    str = realloc(str, space);
    if (i > 2) {
        break;
    }
}

for (i = 0; i < n; i++) {
    printf("%c", str[i]);
}
}

However, when I tried to print out the string from insert(), let's say I entered 'a' to append to the first element of dynamic array with a size of 5, the result that I am getting is abcd=

I referenced from the stackoverflow thread and I not sure how to fix this. Thanks in advance.

解决方案

Here is the code - with the contract that the caller does the free bit! The caller calls it with insert(&str, input, n)

void insert(char **str, char input, int n) {

char* temp = *str;
int i;

*str = realloc(*str, n + 2); /* realloc first */

if(!*str) /* realloc failed */
{
    fputs("realloc failed", stderr);
    free(temp); /* Free the previously malloc-ed memory */
    exit(-1); /* Exit the program */
}

for (i = n; i >= 0; i--) {
    (*str)[i + 1] = (*str)[i]; /* Move all characters up */
}

(*str)[0] = input; /* Insert the new character */

printf("%s", *str); /* Print the new string */
}

Sorry about the formatting. That is left to the reader. I have not checked the algorithm but this does not leak memory

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