调用函数时char array [100]和char * array之间的区别? [英] Difference between char array[100] and char *array when calling functions?

问题描述

我想知道为什么此代码在char tab[100]上可以正常使用，但是如果我使用char *tab则不起作用? fgets函数将char*数组作为参数正确吗?

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);
    }
}

推荐答案

对于无效"，您可能表示您报告了一些严重错误，例如分段错误.

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

char tab[100]和char *tab之间的区别在于，第一个已分配存储，第二个未分配存储.当您使用数组作为参数调用函数时，编译器会将指针传递给数组的第一个元素，因此对于被调用的函数，无论是使用数组参数还是使用数组参数，都看不到区别.带有指针参数.

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.

因此，要让您的程序与char *tab;一起使用，必须首先为该指针分配存储空间，例如与char *tab=malloc(100);一起使用.既然已分配了有效的存储空间(并且指针现在指向它)，则可以调用函数将此tab作为参数.

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.

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