c - 如何纠正我的 C 程序中的段错误

Question

我很难调试为 knapSack 编写的以下程序

#include <stdio.h>
#include <stdlib.h>
#include "timer.h"


#define    MAX(x,y)   ((x)>(y) ? (x) : (y))
#define    table(i,j)    table[(i)*(C+1)+(j)]

int main(int argc, char **argv) {

   FILE   *fp;
   long    N, C, opt;                   // # of objects, capacity 
   int    *weights, *profits, *table, *solution;     // weights and profits
   int    verbose;

  // Temp variables
   long    i, j, count, size, size1, ii, jj;

   // Time
   double time;

   // Read input file:
   //  first line:  # of objects, knapsack capacity, 
   //  next lines:   weight and profit  of next object (1 object per line)
   if ( argc > 1 ) {
      fp = fopen(argv[1], "r"); 
      if ( fp == NULL) {


        printf("[ERROR] : Failed to read file named '%s'.\n", argv[1]);
         exit(1);
      }
   } else {
      printf("USAGE : %s [filename].\n", argv[0]);
      exit(1);
   }

   if (argc > 2) verbose = 1; else verbose = 0;

   fscanf(fp, "%ld %ld", &N, &C);
   printf("The number of objects is %ld, and the capacity is %ld.\n", N, C);

   size    = N * sizeof(int);
   size1   = C * sizeof(int);
   weights = (int *)malloc(size);
   profits = (int *)malloc(size);
   table   = (int *)malloc(size*size1);
   solution= (int *)malloc(size);

   if ( weights == NULL || profits == NULL ) {
      printf("[ERROR] : Failed to allocate memory for weights/profits.\n");
      exit(1);
   }

  for ( i=0 ; i < N ; i++ ) {
      count = fscanf(fp, "%d %d", &(weights[i]), &(profits[i]));
      if ( count != 2 ) {
         printf("[ERROR] : Input file is not well formatted.\n");
         exit(1);
      }
   }

   fclose(fp);



   initialize_timer ();
   start_timer();
   // Solve for the optimal profit (create the table)
    for(j=0; j<=C; j++) {
        table(0,j)=0;
    }

    for(ii=1;ii<=N;ii++) {
        for(jj=0; jj<=C; jj++) {    
            if(weights[ii-1]>jj) {
                table(ii,jj)=table(ii-1,jj);
            }
            else {
                table(ii,jj)=MAX(table(ii-1,jj),(profits[ii-1]+table(ii-1,jj-weights[ii-1])));
            }
            }
        }
opt=table(N,C);
   // We only time the creation of the table

  stop_timer();
   time = elapsed_time ();

   printf("The optimal profit is %ld Time taken : %lf.\n",opt,time);


 // End of "Solve for the optimal profit"


// Find the solution (choice vector) by backtracking through the table


      printf("Solution vector is: \n");
j=C;
      for(i=N;i>0;i--) {
    if(table(i,j)==table(i-1,j)) {
        //printf("Object %d not picked", i);
        solution[i-1]=0;
    }   
    else {
        //printf("Object %d picked", i);
        j=j-weights[i-1];
        solution[i-1]=1;
    }
      }
    for(i=0; i<N; i++) {
        printf("%d ",solution[i]);
    }
      if (verbose) {

    // print the solution vector 
      }

   return 0;
}

对于小输入，代码运行良好。但是对于 N=1200 和 C= 38400000 或 C 的任何其他大输入，代码显示段错误。以下是 Valgrind 的输出:

The number of objects is 1200, and the capacity is 38400000.
==2297== Invalid write of size 4
==2297==    at 0x400A4E: main (knap1.c:73)
==2297==  Address 0x8 is not stack'd, malloc'd or (recently) free'd
==2297== 
==2297== 
==2297== Process terminating with default action of signal 11 (SIGSEGV)
==2297==  Access not within mapped region at address 0x8
==2297==    at 0x400A4E: main (knap1.c:73)
==2297==  If you believe this happened as a result of a stack
==2297==  overflow in your program's main thread (unlikely but
==2297==  possible), you can try to increase the size of the
==2297==  main thread stack using the --main-stacksize= flag.
==2297==  The main thread stack size used in this run was 8388608.
==2297== 
==2297== HEAP SUMMARY:
==2297==     in use at exit: 14,400 bytes in 3 blocks
==2297==   total heap usage: 4 allocs, 1 frees, 14,968 bytes allocated
==2297== 
==2297== LEAK SUMMARY:
==2297==    definitely lost: 0 bytes in 0 blocks
==2297==    indirectly lost: 0 bytes in 0 blocks
==2297==      possibly lost: 0 bytes in 0 blocks
==2297==    still reachable: 14,400 bytes in 3 blocks
==2297==         suppressed: 0 bytes in 0 blocks
==2297== Rerun with --leak-check=full to see details of leaked memory
==2297== 
==2297== For counts of detected and suppressed errors, rerun with: -v
==2297== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 2 from 2)
Segmentation fault

这是使用不同的输入文件(k10.txt 到 k1200.txt)运行时有关局部值(从 gdb 获取)的信息:

  for files with which I got correct output until it exceeded that fix. no. of byte value 
  fp = 0x0 N = 4131212846 C = 140737488347792 opt = 4294967295 weights =

  0x0 profits = 0x36dd221168 table = 0x7ffff7ffc6b8 solution =
  0x36dd8101c0 verbose = 0 i = 0 j = 140737488348128 count = 2 size =
  4198301 size1 = 1 ii = 4196160 jj = 4198224 time =
  2.0728237613194911e-317

  for k1200.txt 
  k1200.txt fp = 0x177b010 N = 1200 C = 38400000 opt = 4294967295
  weights = 0x177b250 profits = 0x177c520 table = 0x8 solution =
  0x7f3cd40008c0 verbose = 0 i = 1200 j = 0 count = 2 size = 4800 size1
  = 153600000 ii = 4196160 jj = 4198224 time = 2.0728237613194911e-317

关于我的代码有什么问题的任何输入？以及如何更正程序以使其永远不会显示段错误？

Answer 1

你在这里要求太多内存:

table   = (int *)malloc(size*size1);

正好 1200 * 38400000 * sizeof (int) * sizeof (int) 大约 74GB 内存(假设 sizeof (int) == 4)。您的计算机实际上无法处理这么大的 block ，因此分配失败，并且当失败时，返回一个 NULL 指针。你应该检查这个条件:

if (table == NULL) {
    fprintf(stderr, "Memory allocation failed :(");
    exit(1);
}

您没有使用 NULL 指针，产生了段错误。

不幸的是，这里没有简单的修复方法。您应该重新考虑该算法，看看您是否真的需要一次这么大的 block ，或者您可以重用一个较小的 block 。

一个小问题是您要求 4 倍于所需的内存(仍然假设 sizeof (int) == 4)。事实上，当您 malloc size * size1 字节时，您将 sizeof (int) 考虑两次，一次是 size = N * sizeof (int) 和一次 size1 = D * sizeof (int)，虽然很明显您想要一个 N * C * sizeof(int)矩阵。

74GB/4 意味着 18.5GB，这仍然太多了:您的操作系统可能能够在虚拟内存中处理它，但是当交换启动时它会变得非常慢。当然，除非您安装了 18+GB 的 RAM .

无论如何，我想您正在使用 table 作为 true/false bool 矩阵。每个元素都可能是一个 32 位的 int，您只使用其中的 1 位。如果使用按位运算将 32 个单元格打包成一个整数，则可以将分配的大小减少 32 倍。它可能会对性能产生影响，但它肯定会将内存占用量减少到您的计算机可以处理的大小。

正如评论中所建议的，您也可以使用 char 或 bool 而不是 int，因为它们通常较小。