Question

我正在尝试更改数组中的值。但我无法弄清楚如何做到这一点而不会产生奇怪的输出。

char *wordsArray[9] = {"word1","word2","word3","word4","word5","word6","word7","word8","word9"};

int *temp;
temp = &wordsArray[randNumber1];

wordsArray[randNumber1] = wordsArray[randNumber2]; //this works
wordsArray[randNumber2] = temp;                    //this does not

我不是指针，所以此刻我不知道我做错了什么欢迎所有帮助。谢谢！

Answer 1

第4行是最大的问题。应该在那里发出编译器警告。通常是类似不兼容的指针。

char *wordsArray[9] = {"word1","word2","word3","word4","word5","word6","word7","word8","word9"};

char *temp; // your array contains char pointers, not integers
temp = wordsArray[randNumber1]; // no need to take the address, you get the array element

wordsArray[randNumber1] = wordsArray[randNumber2];
wordsArray[randNumber2] = temp;                    // now this will work

作为一项学习练习：试着忘记指针：

typedef char * my_string_type;

my_string_type wordsArray[9] = {"word1","word2","word3","word4","word5","word6","word7","word8","word9"};

my_string_type temp = wordsArray[randNumber1];
wordsArray[randNumber1] = wordsArray[randNumber2];
wordsArray[randNumber2] = temp;

Answer 2

int *temp;应为char* temp

您正试图将int存储在char数组中。

Answer 3

int * temp;

表示temp是int *。而且int * temp;与int * temp相同;您可以认为temp是一个int *，即指向int的指针，* temp是一个int。

如果要重现

wordsArray[randNumber1] = wordsArray[randNumber2];

使用temp的行为：

temp = &wordsArray[randNumber3];

请记住，wordsArray [randNumber3]是一个int而＆amp; wordsArray [randNumber3]是指向int的指针。 * temp是一个int。从而你必须输入

wordsArray[randNumber2] = *temp;

想想*和＆amp;作为另一个的倒数，你正在做这样的事情：

wordsArray[randNumber1] = *&wordsArray[randNumber3];

足够令人信服？

更改PointerArray的值

3 个答案: