Question

我在strncat标准库中的string.h函数中看到了一些奇怪的行为，并希望了解一些情况。

我的问题的关键在于我创建的一个名为readLine的函数，其目的是将文件的行作为char *字符串返回，而不使用尾随的换行符终结符。该功能如下所示：

char * readLine(FILE * fp) {
    char * chunk = NULL;
    char * line = NULL;
    int count = 0;

    // iterate through chunks of a line until we reach the end (or an error)
    while (fgets((chunk = malloc(sizeof(char) * BUFFER_SIZE)), BUFFER_SIZE, fp) != NULL) {

        // realloc on a null pointer works like malloc  
        line = realloc(line, ++count * BUFFER_SIZE * sizeof(char));    

        printf("chunk's contents: %s\n", chunk);

        // does chunk contain the end of a line?
        if(strchr(chunk, '\n') == NULL) {   
            // concatenate string parts and continue loop
            strncat(line, chunk, strlen(chunk) + 1);        
            free(chunk);

        } else {
            // we want to return a \0 terminated string without the \n
            // expected position of \n in chunk is ({length of chunk}-1)
            chunk[(strlen(chunk) - 1)] = '\0';

            // concatenate string parts
            strncat(line, chunk, strlen(chunk) + 1);
            printf("readLine line:    %s\n", line);
            free(chunk);

            break;        
        }        
    }
    return line;
}

我在主循环中调用它，看起来像这样：

FILE * fp = NULL;

if ((fp = fopen(FILE_PATH, "r")) != NULL) {
    char * line = NULL;

    while ((line = readLine(fp)) != NULL) {
        printf("main line:        %s\n\n", line);
        free(line);
    }

    fclose(fp);
}

现在奇怪的行为出现在我对#define BUFFER_SIZE 1000的定义中。设置如下，我得到以下输出（这不是我想要的）：

chunk's contents: I am on line 1
readLine line:    I am on line 1
main line:        I am on line 1

chunk's contents: Over here I am on line 2
readLine line:    I am on line 1Over here I am on line 2
main line:        I am on line 1Over here I am on line 2

chunk's contents: Line 3 here
readLine line:    I am on line 1Over here I am on line 2Line 3 here
main line:        I am on line 1Over here I am on line 2Line 3 here

chunk's contents: Look out for 4
readLine line:    I am on line 1Over here I am on line 2Line 3 hereLook out for 4
main line:        I am on line 1Over here I am on line 2Line 3 hereLook out for 4

chunk's contents: Johnny 5 alive!
readLine line:    I am on line 1Over here I am on line 2Line 3 hereLook out for 4Johnny 5 alive!
main line:        I am on line 1Over here I am on line 2Line 3 hereLook out for 4Johnny 5 alive!

但是如果我将该定义更改为#define BUFFER_SIZE 20之类的内容，我会得到我想要的输出：

chunk's contents: I am on line 1

readLine line:    I am on line 1
main line:        I am on line 1

chunk's contents: Over here I am on l
chunk's contents: ine 2

readLine line:    Over here I am on line 2
main line:        Over here I am on line 2

chunk's contents: Line 3 here

readLine line:    Line 3 here
main line:        Line 3 here

chunk's contents: Look out for 4

readLine line:    Look out for 4
main line:        Look out for 4

chunk's contents: Johnny 5 alive!

readLine line:    Johnny 5 alive!
main line:        Johnny 5 alive!

我认为我将问题缩小到strncat(line, chunk, strlen(chunk) + 1);行。当line足够高时，我不明白为什么包含前面的BUFFER_SIZE s。

Answer 1

line = realloc(line, ++count * BUFFER_SIZE * sizeof(char));

不会初始化已分配的内存。因此，如果realloc中的第一个readLine为您提供了上一次调用所获得的内存块 - 并非不可能，那么您可能还有旧内容。

无论如何，对于未初始化的内存，第一个strncat可能会调用未定义的行为，因为分配的内存中不需要0字节。

在进入循环之前将缓冲区分配给line，并将0写入第一个字节。

另外，请勿使用

line = realloc(line, ++count * BUFFER_SIZE * sizeof(char));

如果realloc失败，则会泄漏内存。你应该检查realloc，

的返回值

char *temp = realloc(line, ++count * BUFFER_SIZE * sizeof(char));
if (temp == NULL) {
  // Oops
} else {
    line = temp;
}

malloc来电中不要chunk到fgets，

while (fgets((chunk = malloc(sizeof(char) * BUFFER_SIZE)), BUFFER_SIZE, fp) != NULL)

如果malloc失败，那也会调用未定义的行为。 malloc并在致电fgets之前检查，

while ((chunk = malloc(sizeof(char) * BUFFER_SIZE)) && fgets(chunk, BUFFER_SIZE, fp) != NULL)

Answer 2

尽管如此，你可以坚持使用realloc并将缓冲区设为零。

Answer 3

你的问题在这里：

    line = realloc(line, ++count * BUFFER_SIZE * sizeof(char));

根据realloc的手册页：

"realloc(3) does not guarantee that the additional memory is also
 zero-filled."

和

"If ptr is NULL, realloc() is identical to a call to malloc() for size bytes."

所以你得到的任何新内存都可能被非零字节填充，这意味着第一次调用它时，第一个字节可能不会有0，这意味着strncat将会附加到分配中的任何垃圾字节。

strncat看起来像是在调用时保留数据吗？

3 个答案: