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我的程序收到信号SIGSEGV,分段错误,请帮忙

[英]my Program received signal SIGSEGV, Segmentation fault, help please

When I try to run this program, it doesn't work.当我尝试运行此程序时,它不起作用。 When I go to debugs, a message pops up "program received signal sigsegv, segmentation fault"当我去调试时,弹出一条消息“程序收到信号sigsegv,分段错误”

i debug the programe with gdb, and this is the messages.我用 gdb 调试程序,这是消息。 Program received signal SIGSEGV, Segmentation fault.程序收到信号 SIGSEGV,分段错误。 0x0000555555555195 in ft_atoi (str=0x555555556004 "111") at ft_atoi.c:33 33 while (ft_isspace(str[i])) 0x0000555555555195 in ft_atoi (str=0x555555556004 "111") at ft_atoi.c:33 33 while (ft_isspace(str[i]))

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

static int ft_isspace(char c)
{
  if (c == '\n' || c == '\f' || c == ' ' ||
      c == '\r' || c == '\v' || c == '\t');
  return (1);
  return (0);
}
 
int ft_atoi(const char *str)
{
  int               i;
  int               n;
  int               signe;

  n = 0;
  i = 0;
  signe = 1;
  while (ft_isspace(str[i]))
    i++;
  if (str[i] == '-' || str[i] == '+')
    {
      if (str[i++] == '-')
    signe = -1;
    }
  while (str[i] >= '0' && (str[i] <= '9') && str[i] != '\0')
    {
      if (((unsigned long)n > 9223372036854775807 / 10) ||
      (((unsigned long)n == 9223372036854775807 / 10) &&
       ((unsigned long)n % 10) > 7))
    return ((signe > 0 ? -1 : 0));
      n = n * 10 + (*(str + (i++)) - '0');
    }
  return (n * signe);
}
int main()
{
  printf("%d", ft_atoi("111"));
  return 0;
}
static int ft_isspace(char c)
{
  if (c == '\n' || c == '\f' || c == ' ' ||
      c == '\r' || c == '\v' || c == '\t');
  return (1);
  return (0);
}

This will always return 1, causing the loop to run off the end.这将始终返回 1,导致循环结束。 Notice the semicolon here:注意这里的分号:

      c == '\r' || c == '\v' || c == '\t'); <-- HERE

Imagine if it looked like this:想象一下,如果它看起来像这样:

      c == '\r' || c == '\v' || c == '\t') j=2;

That would cause the if statement to decide whether j=2 is executed.这将导致if语句决定是否执行j=2 And the return on the next line would execute regardless of the results of the if .无论if的结果如何,下一行的return都会执行。

Well your code does that same thing, except instead of j=2 , you have an empty statement.那么你的代码做同样的事情,除了j=2 ,你有一个空语句。

The syntax is: if <expression> <statement>语法是: if <expression> <statement>

Your expression ends with that last close parenthesis.您的表达式以最后一个右括号结束。 So your controlled statement is just the ;所以你的受控语句就是; . .

Just remove that semicolon so the return is the statement that is controlled by the if and that appears immediately after the expression.只需删除该分号,这样return的语句就是由if控制的并且紧接在表达式之后的语句。

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