如果我不在函数中使用 printf 或动态分配，我将无法调用函数。为什么？

Question

First issue : If I don't use "printf" function, I can't call fucntion.

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

char *m_strncpy(char *str1, int count) {
    int i;
    char *new_str;
    for (i = 0; i < count; i++) {
        new_str[i] = str1[i];
    }
    return new_str;
}

int main(void)
{

    char str[80];
    char *sp, e;
    int count;

    printf("Enter a string : ");
    fgets(str, sizeof(str) - 1, stdin);
    printf("The number of character : ");
    scanf("%d", &count);
    //printf("Input complete\n");
    sp = m_strncpy(str, count);
    printf("Cut string is %s\n", sp);
    return 0;
}

如果我不使用 printf("Input complete\n"); 则不会调用 m_strncpy 函数。

Second issue : If I don't use dynamic allocation in m_strncpy function, I can't call function.

Visual Studio 2017 不允许取消初始化。但是Repl编译器允许。

所以当我不初始化char *new_str时，是无法调用的。为什么？？

Answer 1

您的原始代码表现出未定义的行为：

char *m_strncpy(char *str1, int count) {
    int i;
    char *new_str;                  // new_str is not initialized, it doesn't
                                    //  point anywhere
    for (i = 0; i < count; i++) {
        new_str[i] = str1[i];       // dereferencing an uninitialized pointer is undefined
                                    // behaviour, anything can happen.
    }
    return new_str;
}

你可能想要这个：

char *m_strncpy(char *str1, int count) {
  int i;
  char *new_str = malloc(count + 1);  // +1 for the NUL terminator

  if (new_str != NULL)
  {    
    for (i = 0; i < count; i++) {       // copy count first chars
      new_str[i] = str1[i];
    }

    new_str[i] = 0;                     // NUL terminate the string
  }

  return new_str;
}

如果我不在函数中使用 printf 或动态分配，我将无法调用函数。为什么？

I can't call function if i don't use printf or dynamic allocation in function. why?

c

printf

fgets