Question

我编写了一个代码，需要从用户指定的数据文件中读取一系列值（这可能是多达10,000行数据），这是在第一个for循环中完成的。在每行值中，其中一个变量被转换为一个字符串，该字符串定义了另一组输入文件（每个包含~20,000行数据），以便在第二个for循环中进行计算。这适用于前几百次迭代...它正确地读取所有内容，并且得到的计算（下面的代码中省略了计算，因为代码生成相同的问题，有或没有它们）是预期的。

问题是，当第一个for循环达到大约600次迭代时，它会停止并产生错误：分段错误（核心转储）

我很确定这是某种内存问题，因为如果我通过省略第二个循环中的4个输入文件来测试代码，并且它能够达到更高迭代次数....

这是代码：

#include<stdio.h>
#include<math.h>
#include<stdlib.h>

int main()

{
    int num, i, j;
    double X, Y;// 
    float Z;
    char the_data[25], Z_name[10], input_1[50], input_2[50], input_3[50], input_4[50], input_5[50];
    double a, b, c, d, e;

    printf("Enter the name of the data file you are using:\n");
    scanf("%24s", &the_data);
    FILE *DATA = fopen(the_data,"r");

    for (j=1; j<800; j++)
        {

            //***************Read data from a file, the variable Z is a floating point number which, when rounded to the nearest decimal place,  
            //********************determines which directory to load model data from for each value of X and Y
            fscanf(DATA, "%lf %lf %f\n", &X, &Y, &Z);

            //round Z to the nearest 1 decimal place
            Z = Z * 10;
            Z = roundf(Z);
            Z = Z / 10;

            //assign the full directory name to Z_name
            sprintf(Z_name, "%.01fdirectory", Z);

            //assign Z_name to input name string for path to the appropriate data file locations
            sprintf(input_1, "./%s/a.txt", Z_name);
            sprintf(input_2, "./%s/b.txt", Z_name);
            sprintf(input_3, "./%s/c.txt", Z_name);
            sprintf(input_4, "./%s/d.txt", Z_name);
            sprintf(input_5, "./%s/e.txt", Z_name);

            //Open the files
            FILE *input1 = fopen(input_1, "r");
            FILE *input2 = fopen(input_2, "r");
            FILE *input3 = fopen(input_3, "r");
            FILE *input4 = fopen(input_4, "r");
            FILE *input5 = fopen(input_5, "r");


            for (i=1; i < 10000; i++)
                {
                    //For a given Z value, read in the corresponding values. Usually these input files have ~20000 values in each so the loop would be set to run until the end of the file
                    fscanf(input1, "%lf", &a);
                    fscanf(input2, "%lf", &b);
                    fscanf(input3, "%lf", &c);
                    fscanf(input4, "%lf", &d);
                    fscanf(input5, "%lf", &e);

                }
            //Test to see how far it gets in loop before giving up due to segmentation fault    
            printf("The iteration number is: %d\n", j); 

        }
    printf("This will print if the program reaches the end of the first loop\n");

}

我很感激处理这个问题的任何提示或指示。谢谢！

Answer 1

您需要检查fopen()的返回值。由于您永远不会关闭输入文件，因此您可能会达到打开文件的限制。然后fopen()返回NULL，当您尝试将其与fscanf()一起使用时，会出现段错误。

       //Open the files
        FILE *input1 = fopen(input_1, "r");
        if (!input1) {
            printf("open input_1 failed\n");
            exit(1);
        }
        FILE *input2 = fopen(input_2, "r");
        if (!input2) {
            printf("open input_2 failed\n");
            exit(1);
        }
        FILE *input3 = fopen(input_3, "r");
        if (!input3) {
            printf("open input_3 failed\n");
            exit(1);
        }
        FILE *input4 = fopen(input_4, "r");
        if (!input4) {
            printf("open input_4 failed\n");
            exit(1);
        }
        FILE *input5 = fopen(input_5, "r");
        if (!input5) {
            printf("open input_5 failed\n");
            exit(1);
        }

        for (i=1; i < 10000; i++)
            {
                //For a given Z value, read in the corresponding values. Usually these input files have ~20000 values in each so the loop would be set to run until the end of the file
                fscanf(input1, "%lf", &a);
                fscanf(input2, "%lf", &b);
                fscanf(input3, "%lf", &c);
                fscanf(input4, "%lf", &d);
                fscanf(input5, "%lf", &e);

            }
        //Test to see how far it gets in loop before giving up due to segmentation fault    
        printf("The iteration number is: %d\n", j); 

        fclose(input1);
        fclose(input2);
        fclose(input3);
        fclose(input4);
        fclose(input5);

嵌套for循环

1 个答案: