从C中的函数内部分配的字符串数组 [英] Allocate a string array from inside a function in C

问题描述

我有扫描文件，并返回一个字符串数组线沿线数的功能，我的功能看起来像这样：

I have a function that scans a file and returns the number of the lines along with the lines in a string array, my function looks like this :

int load_lines(char* _file, char** _array){
    FILE *infile;
    char line_buffer[BUFSIZ];
    char line_number;

    infile = fopen(_file, "r");

    ine_number = 0;
    while (fgets(line_buffer, sizeof(line_buffer), infile)) 
        ++line_number;

    fclose(infile);
    _array = malloc (line_number * sizeof(char*));
    infile = fopen(_file, "r");
    line_number = 0;

    while (fgets(line_buffer, sizeof(line_buffer), infile)) {
        _array[line_number] = malloc(strlen(line_buffer) + 1);
        strcpy(_array[line_number], line_buffer);

        //a printf on _array[line_number] works fine here
        ++line_number;
    }

    return line_number;
}

当我把它称为是这样的：

When I call it like this:

char** _array;
line_number = load_lines(inname, _array);

_array[0];

我得到一个分段错误，因为该数组似乎对函数的返回后没有分配。

I get a segmentation fault since the array seems to be not allocated after the return of the function.

推荐答案

当你传递一个参数成函数，该函数的总是的工作是在这个参数的副本。

When you pass an argument into a function, the function always works on a copy of that argument.

所以你的情况， load_lines 正在 _array 的副本。原来 _array 不被修改：

So in your case, load_lines is working on a copy of _array. The original _array is not modified:

char** _array = NULL;
printf("%p\n", _array); // Prints "0x0000"
line_number = load_lines(inname, _array);
printf("%p\n", _array); // Prints "0x0000"

要修改 _array ，你需要一个指针传递给它：

To modify _array, you need to pass a pointer to it:

int load_lines(char* _file, char*** _array){
    ...
    (*array) = malloc (line_number * sizeof(char*));
    ...
    (*array)[line_number] = malloc(strlen(line_buffer) + 1);
}

char** _array = NULL;
line_number = load_lines(inname, &_array);

[但是，任何时候你发现自己需要三指针（即 *** ），现在是时候重新考虑你的架构。的]

[However, any time you find yourself needing a triple pointer (i.e. ***), it's time to reconsider your architecture.]

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