Question

我需要帮助；这是示例代码：

#include <stdio.h>
const int n = 4;
const int m = 4;
void display(int arr[n][m]){

   for(int i=0; i<n; i++) {
      for(int j=0;j<m;j++) {
         printf("%d ", arr[i][j]);
      }
    printf("\n");
   }
}

int *constrast(int arr[n][m])
{
   int temp[n][m];
   int max =  25;
   int min =  10;
   int uBound =  255;
   int lBound =  0;
   for(int i=0; i<n; i++) {
      for(int j=0;j<m;j++) {
         temp[i][j] = ((arr[i][j] - max) *((int)(uBound-lBound)/(max-min))+uBound;
      }

   }

   return temp;
}

int main()
{
int disp[4][4] = {
    {10, 11, 12, 13},
    {14, 15, 16, 17},
    {18, 19, 20, 21},
    {22, 23, 24, 25}
};

printf("Image Before Stretching:\n");
display(disp);

printf("Image After Stretching:\n");
display(constrast(disp));
return 0;
}

这是我尝试编译后收到的错误消息：

contrast.c: In function ‘display’:
contrast.c:6:4: error: ‘for’ loop initial declarations are only allowed in C99 mode
    for(int i=0; i<n; i++) {
    ^
contrast.c:6:4: note: use option -std=c99 or -std=gnu99 to compile your code
contrast.c:7:7: error: ‘for’ loop initial declarations are only allowed in C99 mode
       for(int j=0;j<m;j++) {
       ^
contrast.c: In function ‘constrast’:
contrast.c:21:4: error: ‘for’ loop initial declarations are only allowed in C99 mode
    for(int i=0; i<n; i++) {
    ^
contrast.c:22:7: error: ‘for’ loop initial declarations are only allowed in C99 mode
       for(int j=0;j<m;j++) {
       ^
contrast.c:23:82: error: expected ‘)’ before ‘;’ token
          temp[i][j] = ((arr[i][j] - max) *((int)(uBound-lBound)/(max-min))+uBound;
                                                                                  ^
contrast.c:24:7: error: expected ‘;’ before ‘}’ token
       }
       ^
contrast.c:28:4: warning: return from incompatible pointer type [enabled by default]
    return temp;
    ^
contrast.c:28:4: warning: function returns address of local variable [-Wreturn-local-addr]
contrast.c: In function ‘main’:
contrast.c:44:1: warning: passing argument 1 of ‘display’ from incompatible pointer type [enabled by default]
 display(constrast(disp));
 ^
contrast.c:4:6: note: expected ‘int (*)[(sizetype)m]’ but argument is of type ‘int *’
 void display(int arr[n][m]){
      ^

Answer 1

Select * 
From  sys.dm_sql_referenced_entities('dbo.v_View_Obs_Table','Object')

除了int *constrast(int arr[n][m]) { int temp[n][m]; int max = 25; int min = 10; int uBound = 255; int lBound = 0; for(int i=0; i<n; i++) { for(int j=0;j<m;j++) { temp[i][j] = ((arr[i][j] - max) *((int)(uBound-lBound)/(max-min))+uBound; } } return temp; }的类型temp与函数int[4][4]的类型不匹配之外，您永远不要返回局部变量的地址，因为当函数结束，并且指向该函数的指针不再有效。

您不能从C语言中的函数返回数组。一种解决方法是将数组包装在int *中或动态分配内存。更好的是：将输出数组作为参数传递。

Answer 2

当您尝试执行行return temp;时，它仅返回指向数组的指针，然后退出该函数。但是，当您的代码进入int *constrast(int arr[n][m])函数时，该数组已分配到系统堆栈中。返回时，它将删除该变量，而您正尝试使用错误的指针。

作为解决方案，您也可以将结果数组指针作为参数传递，或使用全局变量。

从C中的函数返回二维数组时出错

2 个答案: