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，代码运行正常。但是对于N = 1200和C = 38400000或C的任何其他大输入，代码显示分段错误。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
当使用不同的输入文件（k10.txt到k1200.txt）运行时，以下是有关locals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？如何更正程序，以便它永远不会显示分段错误？
答案 0 :(得分：2)
你在这里要求太多记忆：
table = (int *)malloc(size*size1);
完全1200 * 38400000 * sizeof (int) * sizeof (int)，大约74GB的内存（假设为sizeof (int) == 4）。您的计算机无法彻底处理这么大的块，因此分配失败，当失败时，返回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，导致分段错误。
不幸的是，这里没有简单的解决方法。你应该重新考虑算法，看看你是否确实需要一次这么大的块，或者你可以重用一个较小的块。
一个小问题是你需要4倍的内存（仍然假设为sizeof (int) == 4）。事实上，当您malloc size * size1个字节时，您将sizeof (int)考虑到帐户两次，一次为size = N * sizeof (int)，一次为size1 = D * sizeof (int)，而很明显你想要一个N * C * sizeof(int)矩阵。
74GB / 4意味着18.5GB仍然太多：你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缓慢。除非你安装了18 + GB的RAM，当然
无论如何，我想你使用table作为真/假布尔矩阵。每个元素可能是32位int，其中您只使用1位。如果使用按位运算将32个单元格打包到一个整数中，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减少到计算机可以处理的大小。
根据评论中的建议，您也可以使用char或bool代替int，因为它们通常较小。
答案 1 :(得分：1)
在N = 1200和C = 38400000时，N * C为46,080,000,000。您使用的是32位还是64位操作系统？在32位你的多头可能会溢出。此外，您可能没有足够的内存用于此计算。
看看你的算法，我发现你可能不需要将表分配为N * C，但只需要2 * C.
for循环仅使用行ii-1更新行ii。所以，一旦你计算完毕，ii，你就不再需要ii-1了。这意味着您可以重新使用第ii-1行的内存来存储ii + 1等。因此，您实际上只需要两行。
更像这样：
table = malloc(2*size1);
...
for(ii=1;ii<=N;ii++) {
iiOut = ii%2;
iiIn = (ii-1)%2;
for(jj=0; jj<=C; jj++) {
if(weights[ii-1]>jj) {
table(iiOut,jj)=table(iiIn,jj);
}
else {
table(iiOut,jj)=MAX(table(iiIn,jj),(profits[ii-1]+table(iiIn,jj-weights[ii-1])));
}
}
}
opt=table(iiOut,C);
答案 2 :(得分：0)
好的，除了dohashi的问题之外还有一些事情。
您应该添加检查以查看以下内存分配是否失败：
if ( table == NULL ) {
printf("[ERROR] : Failed to allocate memory for calculation table.\n");
exit(1);
}
if ( solution == NULL) {
printf("[ERROR] : Failed to allocate memory for solution.\n");
exit(1);
}
如果你没有足够的内存来分配这些，现在你就知道了。
接下来，我注意到用于索引到2d表的宏神秘地添加了一个未分配的额外列：
#define table(i,j) table[(i)*(C+1)+(j)]
看到那个“（C + 1）”吗？它说该表的实际大小为N *（C + 1）。接下来，您稍后通过表从1到N和1到C
进行索引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);
宏的大小将table视为大小为N *（C + 1），这实际上要求表格大小为（N + 1）*（C + 2）。
我认为这里至少有一个问题是有人从FORTRAN翻译了这个代码而没有考虑到C中的数组是从零开始而不是从一开始的事实。例如，请参阅here。