如何将 char 连接到常量 char*？

发布于 2024-11-30 17:15:00 字数 399 浏览 2 评论 0原文

我有一个函数连接两个常量 char* 并返回结果。我想做的是将 char 连接到常量 char* 例如

char *command = "nest";
char *halloween =  join("hallowee", command[0]);   //this gives an error

char *join(const char* s1,  const char* s2)
{
    char* result = malloc(strlen(s1) + strlen(s2) + 1);

    if (result)
    {
            strcpy(result, s1);
            strcat(result, s2);
    }

    return result;
}

原文

I have a function that joins two constant char* and returns the result. What I want to do though is join a char to a constant char* eg

char *command = "nest";
char *halloween =  join("hallowee", command[0]);   //this gives an error

char *join(const char* s1,  const char* s2)
{
    char* result = malloc(strlen(s1) + strlen(s2) + 1);

    if (result)
    {
            strcpy(result, s1);
            strcat(result, s2);
    }

    return result;
}

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黄昏下泛黄的笔记 2024-12-07 17:15:00

您编写的函数需要两个 C 字符串（即两个 const char * 变量）。在这里，您的第二个参数是 command[0] ，它不是一个指针 (const char *)，而是一个简单的“n”字符 (const char) >）。然而，该函数认为您传递的值是一个指针，并尝试在由字母“n”的 ASCII 值给出的内存地址中查找字符串，这会导致麻烦。

编辑：要使其工作，您必须更改 join 函数：

char *join(const char* s1,  const char c)
{
    int len = strlen(s1);
    char* result = malloc(len + 2);

    if (result)
    {
            strcpy(result, s1);
            result[len] = c;         //add the extra character
            result[len+1] = '\0';    //terminate the string
    }

    return result;
}

The function you wrote requires two C-strings (i.e. two const char * variables). Here, your second argument is command[0] which is not a pointer (const char *) but a simple 'n' character (const char). The function, however, believes that the value you passed is a pointer and tries to look for the string in memory adress given by the ASCII value of the letter 'n', which causes the trouble.

EDIT: To make it work, you would have to change the join function:

char *join(const char* s1,  const char c)
{
    int len = strlen(s1);
    char* result = malloc(len + 2);

    if (result)
    {
            strcpy(result, s1);
            result[len] = c;         //add the extra character
            result[len+1] = '\0';    //terminate the string
    }

    return result;
}

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染火枫林 2024-12-07 17:15:00

如果您希望加入单个字符，则必须编写一个单独的函数，该函数从 s2 获取要追加的字符数量。

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太傻旳人生 2024-12-07 17:15:00

最好的方法是创建一个新函数，允许将单个字符添加到字符串中。
但是，如果您出于某种原因想按原样使用 join() 函数，您也可以按以下步骤操作：

char *command = "nest";
char *buffer  = " "; // one space and an implicit trailing '\0'
char *halloween;

*buffer = command[0];
halloween = join("hallowee", buffer);

The best is to create a new function that allows adding a single char to a string.
But if you would like to use the join() function as it is for some reason, you can also proceed as follows:

char *command = "nest";
char *buffer  = " "; // one space and an implicit trailing '\0'
char *halloween;

*buffer = command[0];
halloween = join("hallowee", buffer);

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