指向指针的地址分配不起作用 [英] Assigment of adress to pointer doesn't work

问题描述

words_ptr [i] =& words [i]不起作用.
它说它不能从"char(* _ w64)[10]转换为" char"

The words_ptr[i] = &words[i] doesn''t work.
It says that it cannot convert from ''char(*_w64)[10] to ''char''

#define NUM_STRINGS 10
#define STR_LEN 10
int main (void) {
    char words[NUM_STRINGS][STR_LEN];
    char input[STR_LEN];
    char *words_ptr[NUM_STRINGS];
    char *temp;
    int i, j;
    int count = 0;
    for (i = 0; i < NUM_STRINGS; ++i) {
        printf("Enter a word (or 0 to quit): ");
        scanf ("%9s", input);
        discard_input();
        if (input[0] == '0') break;
        strncpy(words[i], input, STR_LEN);
        words_ptr[i] = &words[i];  //mistake
        ++count;
    }

推荐答案

您不需要&"号:words是一个二维数组，因此其行为很像指针数组.

You do not need an ampersand: words is a two-dimensional array, so it behaves a lot like an array of pointers.

正确！您不能这样分配.
但是您可以在分配指针之前先对其进行强制转换.

True! You can''t assign it that way.
But you can cast the pointer before assigning it.

words_ptr[i] = (char*)&words[i]

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