C 奇怪的增量行为与 char 指针地址

Question

我对 char 指针有一个奇怪的行为。根据我对指针的了解，我应该能够通过向 char 指针的每个指向字符添加一个来移动它，因为 char 是一个字节。然而，情况似乎并非如此，使用增量运算符 +=、++，甚至将 char 指针设置为等于自身加一。这些似乎都没有像人们想象的那样影响指针。如果我只是将一个数字或一个变量添加到我的 char 指针，它会像人们期望的那样完美地工作。

这不起作用:

void getNextWord(FILE * pFile)
{
    char * w = (char *)malloc(MAX_WORD_SIZE * sizeof(char *));
    char c;
    while(c != ' ')
    {
        c = fgetc(pFile);
        if(c != ' ')
        {
            *(w++) = c;
        }
    }
    *(w++) = '\0';
    printf("%s",w);
}

这确实有效:

void getNextWord(FILE * pFile)
{
    char * w = (char *)malloc(MAX_WORD_SIZE * sizeof(char *));
    int i = 0;
    char c;
    while(c != ' ')
    {
        c = fgetc(pFile);
        if(c != ' ')
        {
            *(w + i) = c;
            i++;
        }
    }
    *(w + i) = '\0';
    printf("%s",w);
}

有人知道为什么会这样吗？

Answer 1

在第一种情况下，每次添加一个字符时都会递增 w，因此它总是指向刚好超出您添加的最后一个字符。当您打印它时，它指向尚未初始化的内存。

char *s = w;  // Save a pointer to the beginning of the string.
while (c != ' ') {
  c = fgetc(pFile);
  if (c != ' ') {
    // Store the character at w, then increment w
    // to point at the next available (unused) location.
    *(w++) = c;
  }
}
// Null-terminate the string, and increment w again.
// Now it points one location beyond the end of the string.
*(w++) = '\0';

// This will print whatever happens to be in the uninitialized memory
// at w. It will continue to print until it encounters a null character
// (or "illegal" memory, at which point it will crash).
printf("%s", w);

// This will work as expected because it prints the characters that
// have been read.
printf("%s", s);

// This will also work because it "resets" w
// to the beginning of the string.
w = s;
printf("%s", w);