程序收到信号 SIGSEGV，分段错误。 # C 语言#

Question

i have a Program received signal SIGSEGV, Segmentation fault. and i really dunno how to fix it, it says thats the problem is the line of "scanf ( " %s ",& t[i].poste);"

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

typedef struct { int d,m,y; } dtype ;

typedef struct { char nom[10]; char prenom[10] ; char poste[00]; dtype date ; } LEM ;

int main()
{
    int n,i,j;

    printf("Liste des employes\n");
    printf("\n nombre d'employes : ");
    scanf("%d",&n);

    LEM t[n];

    for(i=1;i<=n;i++)
    {
        printf("\nemploye number %d :",i);

        printf("\n nom : ");
        scanf("%s",&t[i].nom);

        printf("\n prenom : ");
        scanf("%s",&t[i].prenom);

        printf("\n poste : ");
        scanf("%s",&t[i].poste);

        printf("\n date de recrutement : ");

        printf("\n day : ");
        scanf("%d",&t[i].date.d);

        printf("\n month : ");
        scanf("%d",&t[i].date.m);

        printf("\n year : ");
        scanf("%d",&t[i].date.y);

    }

    for(i=1;i<=n;i++)
    {
        printf("nom : %s \t",t[i].nom);
        printf("prenom : %s \t",t[i].prenom);
        printf("poste : %s \t",t[i].poste);
        printf("date de rec : %d/%d/%d \t",t[i].date.d , t[i].date.m ,t[i].date.y);

    }
}

Answer 1

If your code is correctly pasted, than you have set poste to be of 0 size (in line typedef struct { char nom[10]; char prenom[10]; char poste[00]; dtype date; } LEM; ). Maybe you meant it to be 10 instead of 00 ?

Answer 2

Your for loops should start at 0 and end when i is less than n since array indexing starts at 0. In your loops first you are skipping the first element and then at the last iteration you are accessing invalid memory and you're getting the segmentation fault.

Another thing to modify is how you handle the scanf -s. When reading something into an array you should check that what you are going to put inside the array fits. What happens if you pass to the program a string that is larger than 9 characters? It may overflow the array and be the cause of another segmentation fault.