字符*（指针）功能 [英] Char * (pointer) function

问题描述

我需要在一个函数一个char *传递，并将它设置为一个CString值。我可以正确设置它作为函数的字符串，但它似乎并没有在那个叫摆在首位的字符*函数的函数正确地打印出来。

  INT l2_read（字符*块，INT长度）
{
    块=的malloc（sizeof的（char）的*长度）;    INT I;
    对于（i = 0; I＆LT;长度;我++）{
       焦炭℃;
       如果（读（和C）℃的）返回（-1）; //这个获取单个字符
          块[I] = C;
    }    的printf（％S，块）; //这个精细打印
    返回1;
}
    //主
    字符*串;
    int值= l2_read（字符串，16）;
    的printf（％S，块）; //打印错误
 

解决方案

在C，一切都是按值传递。一般的规则要记住的是，你不能改变传递给函数的参数值。如果你想传递的东西需要改变，你需要一个指针传递给它。

所以，在你的功能，你想改变块。 块是的char * 。为了能够修改的char *值，你需要传递一个指向，即的char ** 。

  INT l2_read（焦** chunkp，INT长度）
{
    INT I;
    * chunkp =的malloc（长* sizeof的** chunkp）;
    如果（* chunkp == NULL）{
        返回-2;
    }
    对于（i = 0; I＆LT;长度;我++）{
        焦炭℃;
        如果（读（和C）℃的）返回-1;
        （* chunkp）[i] = C;
    }
    的printf（％S，* chunkp）;
    返回1;
}
 

，然后在的main（）：

 的char *字符串;
 int值= l2_read（安培;串，16）;
 如果（价值== 1）{
     的printf（％S，字符串）; / *纠正错字* /
     免费（字符串）; / *调用者调用free（）* /
 }否则如果（价值== -2）{
    / * malloc的失败，处理错误* /
 }其他{
    / *读取失败* /
    免费（字符串）;
 }
 

传递按值用C之所以与strtol（），关于strtod（）等。 ，需要的char ** endptr 参数，而不是的char * endptr ＆MDASH;他们希望能够将的char * 值，以第一个无效字符的地址，他们可以影响一个的char * 在来电者是唯一的出路获得一个指向它，即获赠的char * 。同样，在你的函数，你希望能够改变的char * 值，这意味着你需要一个指针的char * 。

希望有所帮助。

I need to pass in a char * in a function and have it set to a cstring value. I can properly set it as a string in the function, but it doesn't seem to print out correctly in the function that called the char * function in the first place.

int l2_read(char *chunk,int length)
{
    chunk = malloc( sizeof(char) * length);

    int i;
    for(i = 0; i < length; i++){
       char c;
       if(read(&c) < 0) return (-1); // this gets a single character
          chunk[i] = c;
    }

    printf("%s",chunk); // this prints fine
    return 1;
}


    // main
    char *string;
    int value = l2_read(string,16);
    printf("%s",chunk); // prints wrong

解决方案

In C, everything is passed by value. A general rule to remember is, you can't change the value of a parameter passed to a function. If you want to pass something that needs to change, you need to pass a pointer to it.

So, in your function, you want to change chunk. chunk is char *. To be able to change the value of the char *, you need to pass a pointer to that, i.e., char **.

int l2_read(char **chunkp, int length)
{
    int i;
    *chunkp = malloc(length * sizeof **chunkp);
    if (*chunkp == NULL) {
        return -2;
    }
    for(i = 0; i < length; i++) {
        char c;
        if (read(&c) < 0) return -1;
        (*chunkp)[i] = c;
    }
    printf("%s", *chunkp);
    return 1;
}

and then in main():

 char *string;
 int value = l2_read(&string, 16);
 if (value == 1) {
     printf("%s", string); /* corrected typo */
     free(string); /* caller has to call free() */
 } else if (value == -2) {
    /* malloc failed, handle error */
 } else {
    /* read failed */
    free(string);
 }

Pass-by-value in C is the reason why strtol(), strtod(), etc., need char **endptr parameter instead of char *endptr—they want to be able to set the char * value to the address of the first invalid char, and the only way they can affect a char * in the caller is to receive a pointer to it, i.e., receive a char *. Similarly, in your function, you want to be able to change a char * value, which means you need a pointer to a char *.

Hope that helps.

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