我为什么会出现分段错误（核心转储） [英] Why am I getting segmantation fault (core dumped)

问题描述

伙计们，这里有什么不对吗？

  #include   <   stdio.h  >  
   int  main（）
 {
  char  * S，I; 
 printf（ 输入字符串：）; 
 scanf（ ％s，s）; 
  for （i =  0 ; s [i]！= '  \0'; ++ i）; 
 printf（ 字符串长度：％d \ n，i）; 
  return   0 ; 
} 

输出：输入一个字符串：jeevan

字符串长度：6

分段错误（核心转储）

解决方案

  char  * s; 

没有为scanf读取分配任何内存 - 它只是声明一个指针 - 所以你需要使用（例如）

 s = malloc（  100 ）; 

为99个字符的字符串提供足够的空间（加上终止的nul）iirc



问题是，如何确保用户输入的字符不超过99个？ - 答案是，使用像这样的scanf，你不能 - 它'危险'



我看过有关scanf的说明符的说法，暗示使用％ms获取scanf为你分配内存，但是，我无法证明这种用法[/ edit]



我似乎记住你也可以使用scanf作为长度说明符，但是，你仍然必须使用malloc或声明s作为char数组，当用户键入你允许的字符数时，它会变得混乱



- 你应该使用更安全的东西，比如fgets（）而不是



'g'

你没有为这个字符串分配任何内存，快速和脏的修复将是：

 #include< stdio.h> 
  int  main（）
 {
  char  s [ 255 ]，I; 
 printf（ 输入字符串：）; 
 scanf（ ％s，s）; 
  for （i =  0 ; s [i]！= '  \0'; ++ i）; 
 printf（ 字符串长度：％d \ n，i）; 
  return   0 ; 
} 

一个不错的读物：字符串溢出with scanf [ ^ ]。

guys, whats wrong here?

#include <stdio.h>
int main()
{
    char *s,i;
    printf("Enter a string: ");
    scanf("%s",s);
    for(i=0; s[i]!='\0'; ++i);
    printf("Length of string: %d \n",i);
    return 0;
}

output:Enter a string: jeevan
Length of string: 6
Segmentation fault (core dumped)

解决方案

char *s;

doesn't allocate any memory for the scanf read - it just declares a pointer - so you need to use (for example)

s = malloc(100);

to get enough space for a 99 character string (plus the terminating nul) iirc

the issue then is, how do you ensure your user doesn't type more than 99 characters ? - and the answer is, using scanf like this, you cant - its 'dangerous'

[edit] I have seen talk of a specifier for scanf, that alludes to using "%ms" gets scanf to allocate the memory for you, but, I cannot attest to this form of usage [/edit]

[edit] I seem to remember that you could also a length specifier with scanf, but, you'd still have to use the malloc or declare s as an array of char, and, it gets messy when the user types over the number of characters you've allowed for [edit]

- you should use something safer, like fgets() instead

'g'

You have not allocate any memory for this string, the quick and dirty fix would be:

#include <stdio.h>
int main()
{
    char s[255],i;
    printf("Enter a string: ");
    scanf("%s",s);
    for(i=0; s[i]!='\0'; ++i);
    printf("Length of string: %d \n",i);
    return 0;
}

A nice reading: "String overflows with scanf"[^].

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