任何人都知道分段错误在哪里？

Question

我是 C 的新手，所以很抱歉，如果它很明显，但在过去的一个半小时里我一直在努力寻找这个分段错误。我一直在注释代码的各个部分以尝试找到它。但是当我修复我认为错误的地方时，分段错误仍然存在。我一直在搜索的代码如下。任何帮助将不胜感激。最初我认为这是计数器的问题，但后来看起来更像是行。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/time.h>
#include <time.h>
#include "utils.h"

int main(int argc, char* argv[])
{
  struct tm *local;
  struct tm *local2;
  time_t start, end;
  time(&start); // read and record clock
  local = localtime(&start);
  struct timeval tim;
gettimeofday(&tim, NULL);
double t1=tim.tv_sec+(tim.tv_usec/1000000.0);

  printf("# Start time and date: %s", asctime(local));
  float refVal;
  float tolVal;
  int vFirst, vSecond;
  vFirst = (strcmp(argv[1], "-v") == 0);
  vSecond = (strcmp(argv[5], "-v") == 0);

 int rct;
  int cct;
  int j;
  float row = 0;
  int r;
  int c;

  int counter=0;
  int refFirst;
  int v = 0;
  scanf("%d %d",&rct,&cct);

  if(vFirst){
    v = 1;
    refFirst = (strcmp(argv[2], "-r")== 0);
    if(refFirst){
      refVal = strtof(argv[3], NULL);
      tolVal = strtof(argv[5], NULL);
    }else{
      tolVal = strtof(argv[3], NULL);
      refVal = strtof(argv[5], NULL);
    }
 }else if(vSecond){
    v = 1;
    refFirst = (strcmp(argv[1], "-r")== 0);
    if(refFirst){
      refVal = strtof(argv[2], NULL);
      tolVal = strtof(argv[4], NULL);
    }else{
      tolVal = strtof(argv[2], NULL);
      refVal = strtof(argv[4], NULL);
    }
  }else{
     refFirst = (strcmp(argv[1], "-r") == 0);
    if(refFirst){
      refVal = strtof(argv[2], NULL);
      tolVal = strtof(argv[4], NULL);
    }else{
      tolVal = strtof(argv[2], NULL);
      refVal = strtof(argv[4], NULL);
    }

    }

float** rows = (float **) malloc(rct * sizeof(float *));
if (rows == 0)
{
fprintf(stderr, "Couldn’t alocate sufficient space.\n");
exit(1);
}
int i;
for (i = 0; i < rct; i++)
{
float* row = (float *) malloc(cct * sizeof(float));
if (row == 0)
{
fprintf(stderr, "Couldn’t alocate sufficient row space.\n");
exit(1);
}
rows[i] = row;
}

  for(i = 0; i < rct; i++)
    {
        for(j = 0; j < cct; j++)
        {
            scanf("%f", &rows[i][j]);
        }
    }
    for(i = 0; i < rct; i++)
      {
        for(j = 0; j < cct; j++)
          {
            if(approxEqual(rows[i][j], refVal, tolVal)){
              counter++;
              if(v != 1){
              fprintf(stdout, "r=%d, c=%d: %.6f\n", i, j, rows[i][j]);
              }
            }
          }
          }


    char found[] = "Found";
    char apprx[] = "approximate matches.";
    printf("%s %d %s\n", found, counter, apprx);
    gettimeofday(&tim, NULL);
double t2=tim.tv_sec+(tim.tv_usec/1000000.0);
printf("%.6lf seconds elapsed\n", t2-t1);

}

Answer 1

解决方案已在评论中确定，但在建议处理 command-line 输入相关错误的方式中，使用 command-line 输入时，最好包含一些用法语句提醒用户（包括您自己）是否在没有正确的 number/type 参数的情况下调用可执行文件。

简单示例：

int main(int argc, char* argv[])
{
     if(argc != 6)
     {
          printf("argc == %d, expected 6.\n"Usage: some.exe <arg1> <arg2> ... <arg_n>\nProgram will now exit.", argc);
          return 0;
     }
     ...

任何人都知道分段错误在哪里？

Anyone Know Where The Segmentation Fault Is In This?

c

segmentation-fault