编译器显示错误消息“从整数生成指针而不进行强制转换". (反转数组).我该如何解决? [英] Compiler shows the error message "makes pointer from integer without cast" (reversing an array). How do i resolve this?

问题描述

代码如下.调用函数reverse_array时，如何将第一个函数参数定义为数组?

The code goes as follows. while calling the function reverse_array, How do I define the first function argument as an array?

#include <stdio.h>

void swap(int* a, int* b);
void reverse_array(int a[], int n) {
  int* b = a + n - 1;
  while (b > a) {
    swap(a, b);
    a = a + 1;
    b = b - 1;
  }
}

void swap(int* ptra, int* ptrb) {
  int t;
  t = *ptra;
  *ptra = *ptrb;
  *ptrb = t;
}

main() {
  int b[5] = { 1,2,3,4,5 };
  reverse_array(b[5], 5);
  int x;
  for (x = 0; x < 5; x++) {
    printf("%d", b[x]);
  }
}

推荐答案

编译器显示错误消息使指针从整数生成而不进行强制转换"(反转数组).我该如何解决?

请勿将int传递到期望指针的位置. :)

Do not pass an int to where a pointer is expected. :)

背景:

在函数定义的上下文中，类似int a[]的内容与int * a相同，因此a是指向int的指针.

In the context of a function definition something like int a[] is the same as int * a, so a is a pointer to int.

您的呼叫者传递了b[5]，这是int-数组b的第六个元素，因此它是int(并且顺便说一句，越界访问，就像在C数组中是，因此b的元素可通过0到4之间的索引来寻址.

You caller passes b[5], which is the 6th element of the int-array b, so it's an int (and BTW an out-of-bound access, as in C array are 0-based so b's element are addressed via indexes ranging from 0 to 4.

要传递"数组，只需给出b.

To "pass" the array just give b.

  reverse_array(b, 5);

然后，数组b将被衰减到其第一个元素的地址.此地址在函数中作为值a接收.

The array b would then be decayed to the address of its 1st element. This address is received within the function as value a.

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