C++ 昨天分段错误测试的代码工作完美，现在它出错了

Question

程序到目前为止计算

ASCII代码中每个字符的出现次数，昨天测试时，它运行良好，即使在我尝试过的最大的情况下也没有错误（小说Huck Finn）。我今天加载了它并再次测试了它，它给了我一个分段错误。我没有更改或添加任何代码。它与昨天完美运行的代码相同。我不明白它今天会如何工作，而不是第二天。运行调试器时，我得到这个：

    Program received signal SIGSEGV, Segmentation fault.
    0x00007ffff7343f01 in getc () from /lib64/libc.so.6

我的代码：

#include <cstdio>
using namespace std;
//==============================================================
//                  initialize
//==============================================================
// initialize(freq) initializes all values in the array pointed
// to by freq to 0.
//==============================================================
void initialize(int* freq)
{
   for(int i = 0; i<=256; i++)
   {
      freq[i]=0;
   } 
}
//==============================================================
//                  getFreq
//==============================================================
// getFreq(inf, freq) counts the number of occurences of each
// character in file inf and stores those numbers into the array
// freq at with the index number being the ASCII code for that 
// character
//==============================================================
void getFreq(FILE* inf, int* freq)
{
   int c = fgetc(inf);
   while(c != EOF)
   {
      freq[c]++;
      c = fgetc(inf);
   }  
}
//==============================================================
//                  showCount
//==============================================================
// showCount(freq) prints the counts of the characters stored in
// freq at with the indexs being the ASCII code for each charater.
// *Does not print characters that do not occur
//==============================================================
void showCount(int* freq)
{
   for(int i = 0; i <= 256; i++)
   {
      if(freq[i]> 0)
      {
         if(i == ' ')
         {
            printf("Spaces: %i n", freq[i]);
         }
         else if(i == 'n')
         {
            printf("Newlines: %i n", freq[i]);
         }
         else if(i == 't')
         {
            printf("Tabs: %i n", freq[i]);
         }
         else if(i == '.')
         {
            printf("Periods: %i n", freq[i]);
         }
         else
         {
            printf("%c's: %i n",i , freq[i]);
         }
      }
   }
}
int main(int argc, char* argv[])
{
   int* freq = new int[256];
   initialize(freq);
   FILE* inf = fopen(argv[1], "r");
   getFreq(inf, freq);
   showCount(freq);
   fclose(inf);
}​

Answer 1

当数组实际上只有 256 个从索引 0 到索引 255 的元素时，您在代码中无处不在地使用 i<=256。代码表现出未定义的行为。昨天工作正常，今天咬我的屁股是一个完美的例子，说明当代码展示 UB 时会发生什么。

Answer 2

您不会通过访问数组外部的元素（从 0-255 的数组的索引 256）来检查 fopen 的返回值和溢出缓冲区。