free() function 导致断点

Question

{
char name_entry[ARRAY_SIZE];//bunu dinamik olarak atamıyorum her seferinde gireceğimiz ismin uzunluğunu sormak pratik olmayacaktır
int isimsayisi;
char** names;
printf("Kac Isim Gireceksiniz:");
scanf("%d", &isimsayisi);
names = (char**)malloc(sizeof(char) * (isimsayisi));
for (int k = 0; k < isimsayisi; k++) {
    printf("\nismi giriniz :");
    scanf("%s", name_entry);
    names[k] = (char*)malloc(strlen(name_entry) + 1);
    if (names[k] == NULL) {
        printf("bellek yetersiz!..\n");
        exit(EXIT_FAILURE);
    }
    strcpy(names[k], name_entry);
}
printf("\nGirilen Isimler : \n");
for (int k = 0; k < isimsayisi; k++) {
    printf("%s \n", names[k]);
}
printf("\n");
for (int i = 0; i < isimsayisi; i++)
free(names[i]);
_getch();
} 

In this code Im try to take names to a 2d string array then printf what have ı took from user then deallocate the memory but ı cant if I print free(names); after the for loop it gives a windows tab error. Its work like that to 2 names if I take more then 2 names its causes a breakpoint Edit:isimsayisi means how much names the users wants to enter

Answer 1

This

names = (char**)malloc(sizeof(char) * (isimsayisi));

should be

names = malloc(sizeof(char *) * (isimsayisi));

1. you don't need to cast malloc() in C (different story for C++).
2. names is char ** ; therefore, when allocating memory, you need to use the sizeof(char *) , which is definitely different than sizeof(char) .

sizeof(char) is the size of a single character and is guaranteed to be 1, while sizeof(char *) is the size of an address and varies based on your machine (8 for a 64-bit CPU).