我在 spbox.c
中有以下代码:
#include <stdbool.h>
#include <stdint.h>
typedef struct {
bool initialized;
uint32_t Spbox[8][64]; // Combined S and P boxes
} spboxState;
static spboxState stateInstance;
uint32_t ** desGetSpbox(void) {
if (!(stateInstance.initialized)) {
// TODO: Compute data to populate Spbox array
stateInstance.initialized = true;
}
return stateInstance.Spbox;
}
我编译它:
clang -c spbox.c
我收到关于不兼容指针返回类型的警告:
spbox.c:16:9: warning: incompatible pointer types returning 'uint32_t [8][64]' from a function with result type 'uint32_t **' (aka 'unsigned int **') [-Wincompatible-pointer-types]
return stateInstance.Spbox;
^~~~~~~~~~~~~~~~~~~
1 warning generated.
如何更改我的代码以使警告消失?这是说 uint32_t **
与 uint32_t [8][64]
不兼容。但是,如果我尝试使后者成为返回类型,则会出现语法错误。
最佳答案
您不能返回数组。但是您可以返回一个指向数组或其第一个元素的指针。
例如
uint32_t ( * desGetSpbox(void) )[8][64] {
if (!(stateInstance.initialized)) {
// TODO: Compute data to populate Spbox array
stateInstance.initialized = true;
}
return &stateInstance.Spbox;
}
或者
uint32_t ( * desGetSpbox(void) )[64] {
if (!(stateInstance.initialized)) {
// TODO: Compute data to populate Spbox array
stateInstance.initialized = true;
}
return stateInstance.Spbox;
}
这是一个演示程序
#include <stdio.h>
#include <stdint.h>
#include <stdbool.h>
#include <inttypes.h>
enum { M = 8, N = 64 };
typedef struct {
bool initialized;
uint32_t Spbox[M][N]; // Combined S and P boxes
} spboxState;
static spboxState stateInstance;
uint32_t ( *desGetSpbox1( void ) )[M][N]
{
stateInstance.Spbox[0][0] = 10;
return &stateInstance.Spbox;
}
uint32_t ( *desGetSpbox2( void ) )[N]
{
stateInstance.Spbox[0][0] = 10;
return stateInstance.Spbox;
}
int main(void)
{
uint32_t ( *p1 )[M][N] = desGetSpbox1();
printf( "%" PRIu32 "\n", ( *p1 )[0][0] );
uint32_t ( *p2 )[N] = desGetSpbox2();
printf( "%" PRIu32 "\n", ( *p2 )[0] );
return 0;
}
程序输出为
10
10
关于c - 如何在 C 中返回静态分配的二维数组?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/66552942/