c - 如何传递对字符串的引用？

Question

我读过的关于 scanf、gets 和 fgets 的所有内容都表明它们是有问题的；带有空白、溢出或复杂性。我正在学习 C 类(class)简介，因为我有足够的 Java 和其他语言编程经验，因此我有信心这样做，所以我决定创建自己的函数来使用 getchar()getchar() 函数。我的代码的相关部分如下:

bool get_string (char prompt[], char* string)
{
    printf(prompt); // Prompt the user

    // Ensure string is allocated to store the null character and reset it if it has been initialized
    do { string = (char*)malloc(sizeof(char)); } while (string == NULL);

    int index = 0;
    char place = getchar();
    while (place != '\n') // Obtain all characters before endl
    {
        string = (char*)realloc(string, sizeof(string) + sizeof(char)); // Create room in the array
        if (string == NULL) return false; // Ensure realloc worked correctly
        string[index++] = place; // Place the new string in the second to last index
        place = getchar();
    }
    string[index] = '\0'; // Append the null character at the end

    return true; // Operation succeeded
}

通过测试和调试，我设法弄清楚:

1. 我的函数在本地符合规范，参数string保存输入的字符串。
2. 我在 main 方法中使用的 char* 指针没有被更改。调用我的输入函数后，该指针的取消引用保持与其初始值相同。

我的印象是，因为我传递了一个指向函数的指针，所以它会将参数视为引用。事实上，这就是我在类里面所教的。任何见解都会有所帮助。

如果出现以下情况，将获得奖励积分:

你可以告诉我为什么它不允许我释放 main 中的 char* 指针。 (也许是因为它尚未通过同一问题分配？)

我还做错了什么，例如多次调用 realloc？

注意:我使用 MSVC C89 编译器并定义 bool、true 和 false 预编译。

Answer 1

I was under the impression that because I was passing a pointer to the function it would treat the parameter as by reference. In fact, this is what I was taught in the class. Any insight can help.

C 中没有引用。您传递的每个参数都是按值。

因此:

void foo(int a) {
  a = 21;
  // a == 21 for the rest of THIS function
}
void bar(void) {
  int x = 42;
  foo(x);
  // x == 42
}

同样适用:

static int john = 21;
static int harry = 42;
void foo(int * a) {
  a = &john;
  // a points to john for the rest of THIS function
}
void bar(void) {
  int * x = &harry;
  foo(x);
  // x still points to harry
}

如果您想通过参数更改指针，那么您需要将指针传递给该指针:

static int john = 21;
static int harry = 42;
void foo(int ** m) {
  *m = &john;
}
void bar(void) {
  int * x = &harry;
  foo(&x); // passing the address of x by value
  // x now points to john
}

What else I am doing wrong, such as calling realloc too many times?

printf(prompt);

安全问题:尝试将 "%s" 等内容作为 prompt 的值。最好使用 puts 或 printf("%s", Prompt)。

do { string = (char*)malloc(sizeof(char)); } while (string == NULL);

这可能是一个无限循环。如果malloc失败，立即再次调用它不会改变任何东西。另外:不要强制转换 malloc 的返回值。此外，sizeof(char)定义等于1。

int index = 0;

对于索引，请使用size_t。

char place = getchar();

getchar 返回 int 是有原因的，即能够检查 EOF，您...

while (place != '\n')

...不，但应该!

string = (char*)realloc(string, sizeof(string) + sizeof(char));

不要转换返回值，sizeof(string) 并没有按照你想象的那样做，它是一个编译时常量(在 64 位上可能是 8系统)。

if (string == NULL) return false;

内存泄漏，因为...

If there is not enough memory, the old memory block is not freed and null pointer is returned.

[Source]

---

这是我在 C 中读取一行的方法:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

char * readline(char const * const prompt) {
  char buffer[10];
  char * string = malloc(1);
  if (string == NULL) {
    return NULL;
  }
  // The accumulated length of the already read input string.
  // This could be computed using strlen, but remembering it
  // in a separate variable is better, performancewise.
  size_t accumLength = 0;
  string[0] = '\0';
  printf("%s", prompt);
  while (fgets(buffer, 10, stdin) != NULL) {
    // To see what has been read in this iteration:
    // printf("READ: |%s|\n", buffer);
    // Compute the length of the new chunk that has been read:
    size_t const newChunkLength = strlen(buffer);
    // Try to enlarge the string so that the new chunk can be appended:
    char * const newString = realloc(string, accumLength + newChunkLength + 1);
    if (newString == NULL) {
      free(string);
      return NULL;
    }
    string = newString;
    // Append the new chunk:
    strcpy(string + accumLength, buffer);
    accumLength += newChunkLength;
    // Done if the last character was a newline character
    assert(accumLength > 0);
    if (string[accumLength - 1] == '\n') {
      // NOTE: Wasting 1 char, possible solution: realloc.
      string[accumLength - 1] = '\0';
      return string;
    }
  }
  // EOF is not an error!
  if (feof(stdin)) {
    return string;
  }
  free(string);
  return NULL;
}


int main(int argc, char ** argv) {
  char const * const input = readline(">");
  printf("---\n%s\n---\n", input);
  return 0;
}