c - for 循环突然停止 C

Question

大家好，我的 for 循环中发生了一件非常奇怪的事情。

当我在这里执行这段代码时:

#include <stdio.h>
#include <string.h>    
char* repeat(char c, int n);

int main(void)
{
    char* input;
    input = repeat('c', 12);

    return 0;
}

char* repeat(char c, int n)
{
    char* out;

    for (int i = 0; i < 12; ++i) //FIX ITERATION
    {
        int len = strlen(out);
        out[len] = c;
        out[len+1] = '\0';
    }

    printf("%s\n", out);
    return out;
}

我得到了预期的输出:

cccccccccccc

但是当我在我的方法中使用传递的 int 时，像这样:

char* repeat(char c, int n)
{
    char* out;

    for (int i = 0; i < n; ++i) //VARIABLE ITERATION
    {
        int len = strlen(out);
        out[len] = c;
        out[len+1] = '\0';
    }

    printf("%s\n", out);
    return out;
}

我只是得到这个作为输出:

cccc

请告诉我我做错了什么。我不知道错误可能是什么？

感谢您的帮助!

Answer 1

这一行就是问题所在

  int len = strlen(out);
  out[len] = c;
  out[len+1] = '\0';

out 未初始化。您没有在此语句中分配内存:

char* out;

所以你正在经历Undefined Behavior .

第 3.4.3 节

1 undefined behavior

behavior, upon use of a nonportable or erroneous program construct or of erroneous data, for which this International Standard imposes no requirements

第 4.1 节:

An lvalue (3.10) of a non-function, non-array type T can be converted to an rvalue. If T is an incomplete type, a program that necessitates this conversion is ill-formed. If the object to which the lvalue refers is not an object of type T and is not an object of a type derived from T, or if the object is uninitialized, a program that necessitates this conversion has undefined behavior. If T is a non-class type, the type of the rvalue is the cv-unqualified version of T. Otherwise, the type of the rvalue is T.

在这两个示例中，您都显示了未定义的结果。

你必须分配内存:

char * out = malloc(sizeof(char)*50); // i have used size 50 - take sufficient what you need
//initialize it 
out[0] = '\0';

确保包含 stdlib.h。

out 现在指向一个可以容纳 50 个 char 的内存块。