Difference between char array[100] and char *array when calling functions?

问题

i'd like to know why this code works fine with char tab[100] but doesn't work if I use char *tab ? fgets function takes a char* array as a parameter right ?

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

 Int Palindrome(char* str, int i, int j);

 int main()
{
    char tab[100];
    printf("Enter your string : \n");
    fgets(tab, 100, stdin); 
    int j = strlen(tab);
    printf("%d\n", Palindrome(tab, 0, j - 2));
    return 0;
}

int Palindrome(char* str, int i, int j)
{
    if (i >= j)
    {
        printf("My word is a Palindrome !\n");
        return printf("<(^w^)>\n");
    }
    else if (str[i] != str[j])
    {
        printf("My word is not a Palindrome !\n");
        return printf("<(X.X)>\n");
    }
    else 
    {
        return Palindrome(str, i + 1, j - 1);
    }
}

回答1:

With "not work" you probably mean you get some serious error reported like a segmentation fault.

The difference between char tab[100] and char *tab is that the first has storage allocated and the second hasn't. When you call a function with an array as a parameter, then the compiler passes a pointer to the first element of the array, so for the function that got called it doesn't see the difference whether it is called with an array-parameter or with a pointer-parameter.

So to let your program work with char *tab; you must first allocate storage to this pointer, such as with char *tab=malloc(100); Now that there is valid storage allocated (and the pointer now points to it), you can call your function with this tab as parameter.

来源：https://stackoverflow.com/questions/45162908/difference-between-char-array100-and-char-array-when-calling-functions

标签

arrays

pointers

recursion

palindrome