Question

我正在使用动态内存分配编写此代码，对于学生记录如图所示，这段代码假设很简单，我显然是以正确的方式在正确的位置分配元素，但是当它打印它们时，它给了我一个“core dumped”错误！怎么了？

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void main()
{
    char **firstname;
    char **lastname;
    float *score;
    int number_of_records,i,j,ctr=1,row=15,col=20;
    /*ctr is to keep track of the student's number (makes it easier to
      the user), it starts with (1)*/

    firstname=malloc(row*sizeof(char*));
    for(i=0;i<row;i++)
    {
        firstname[i]=malloc((col+1)*sizeof(char));
    }
     lastname=malloc(row*sizeof(char*));
    for(i=0;i<row;i++)
    {
        lastname[i]=malloc((col+1)*sizeof(char));
    }
    printf("\nPlease indicate number of records you want to enter (min 2, max 15): ");
    scanf("%d",&number_of_records);
    score=malloc(row*sizeof(float));

    printf("\nPlease input records of students\n(enter a new line after"
           "each record), with following format:\nfirst name last name score ");
    for (i=0;i<number_of_records;i++)
    {
        printf("\nEnter record for student %d : ",ctr);
        scanf("%s %s %f",firstname[i],lastname[i],score[i]);

        ctr++; /*ctr is to keep track of student number
                 (makes it easy to the user) */

    }
     for (i=0;i<number_of_records;i++)
    {

        printf("%s %s %f\n",firstname[i],lastname[i],score[i]);
    }
}

Answer 1

请更改此内容：

for (i=0;i<number_of_records;i++)
    {
        printf("\nEnter record for student %d : ",ctr);
        scanf("%s %s %f",&firstname[i],&lastname[i],&score[i]);

        ctr++; /*ctr is to keep track of student number (makes it easy to the       user) */
    }

到这个

for (i=0;i<number_of_records;i++)
    {
        printf("\nEnter record for student %d : ",ctr);
        scanf("%s %s %f",firstname[i],lastname[i],&score[i]);

        ctr++; /*ctr is to keep track of student number (makes it easy to the       user) */
    }

它会起作用。原因是firstname[i]和lastname[i]已经pointers;您不需要使用&运算符。

P.S。此外，由于您使用的是C，请不要从malloc或realloc转换返回值。它将自动完成。

如何在C中实现动态数组？

1 个答案: