C语言 C将int数组指针作为参数传递给函数
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C pass int array pointer as parameter into a function
提问by stergosz
I want to pass the B int array pointer into func function and be able to change it from there and then view the changes in main function
我想将 B int 数组指针传递给 func 函数，并能够从那里更改它，然后查看 main 函数中的更改
#include <stdio.h>
int func(int *B[10]){
}
int main(void){
int *B[10];
func(&B);
return 0;
}
the above code gives me some errors:
上面的代码给了我一些错误：
In function 'main':|
warning: passing argument 1 of 'func' from incompatible pointer type [enabled by default]|
note: expected 'int **' but argument is of type 'int * (*)[10]'|
EDIT: new code:
编辑：新代码：
#include <stdio.h>
int func(int *B){
*B[0] = 5;
}
int main(void){
int B[10] = {NULL};
printf("b[0] = %d\n\n", B[0]);
func(B);
printf("b[0] = %d\n\n", B[0]);
return 0;
}
now i get these errors:
现在我收到这些错误：
||In function 'func':|
|4|error: invalid type argument of unary '*' (have 'int')|
||In function 'main':|
|9|warning: initialization makes integer from pointer without a cast [enabled by default]|
|9|warning: (near initialization for 'B[0]') [enabled by default]|
||=== Build finished: 1 errors, 2 warnings ===|
回答by Daniel Fischer
In your new code,
在您的新代码中，
int func(int *B){
*B[0] = 5;
}
Bis a pointer to int, thus B[0]is an int, and you can't dereference an int. Just remove the *,
B是指向 的指针int，因此B[0]是int，并且您不能取消引用int。只需删除*,
int func(int *B){
B[0] = 5;
}
and it works.
它有效。
In the initialisation
在初始化
int B[10] = {NULL};
you are initialising anintwith a void*(NULL). Since there is a valid conversion from void*to int, that works, but it is not quite kosher, because the conversion is implementation defined, and usually indicates a mistake by the programmer, hence the compiler warns about it.
您正在int使用void*( NULL)初始化 an 。由于有一个从void*to的有效转换int，这是可行的，但它并不完全符合，因为转换是实现定义的，并且通常表明程序员有错误，因此编译器会对此发出警告。
int B[10] = {0};
is the proper way to 0-initialise an int[10].
是 0 初始化 an 的正确方法int[10]。
回答by Alberto Bonsanto
Maybe you were trying to do this?
也许你试图这样做？
#include <stdio.h>
int func(int * B){
/* B + OFFSET = 5 () You are pointing to the same region as B[OFFSET] */
*(B + 2) = 5;
}
int main(void) {
int B[10];
func(B);
/* Let's say you edited only 2 and you want to show it. */
printf("b[0] = %d\n\n", B[2]);
return 0;
}
回答by greggo
If you actually want to pass an array pointer, it's
如果你真的想传递一个数组指针，它是
#include <stdio.h>
void func(int (*B)[10]){ // ptr to array of 10 ints.
(*B)[0] = 5; // note, *B[0] means *(B[0])
//B[0][0] = 5; // same, but could be misleading here; see below.
}
int main(void){
int B[10] = {0}; // not NULL, which is for pointers.
printf("b[0] = %d\n\n", B[0]);
func(&B); // &B is ptr to arry of 10 ints.
printf("b[0] = %d\n\n", B[0]);
return 0;
}
But as mentioned in other answers, it's not that common to do this. Usually a pointer-to-array is passed only when you want to pass a 2d array, where it suddenly looks a lot clearer, as below. A 2D array is actually passed as a pointer to its first row.
但正如其他答案中提到的，这样做并不常见。通常只有当你想传递一个二维数组时才传递一个指向数组的指针，它突然看起来清晰了很多，如下所示。二维数组实际上是作为指向其第一行的指针传递的。
void func( int B[5][10] ) // this func is actually the same as the one above!
{
B[0][0] = 5;
}
int main(void){
int Ar2D[5][10];
func(Ar2D); // same as func( &Ar2D[0] )
}
The parameter of func may be declared as int B[5][10], int B[][10], int (*B)[10], all are equivalent as parameter types.
func 的参数可以声明为int B[5][10], int B[][10], int (*B)[10]，都等价于参数类型。
Addendum: you can return a pointer-to-array from a function, but the syntax to declare the function is very awkward, the [10] part of the type has to go after the parameter list:
附录：你可以从函数返回一个指向数组的指针，但是声明函数的语法很笨拙，类型的 [10] 部分必须在参数列表之后：
int MyArr[5][10];
int MyRow[10];
int (*select_myarr_row( int i ))[10] { // yes, really
return (i>=0 && i<5)? &MyArr[i] : &MyRow;
}
This is usually done as below, to avoid eyestrain:
这通常如下进行，以避免眼睛疲劳：
typedef int (*pa10int)[10];
pa10int select_myarr_row( int i ) {
return (i>=0 && i<5)? &MyArr[i] : &MyRow;
}
回答by Sunil Bojanapally
In new code assignment should be,
在新的代码分配中应该是，
B[0] = 5
In func(B), you are just passing address of the pointer which is pointing to array B. You can do change in func() as B[i] or *(B + i). Where i is the index of the array.
在 func(B) 中，您只是传递指向数组 B 的指针的地址。您可以将 func() 更改为 B[i] 或 *(B + i)。其中 i 是数组的索引。
In the first code the declaration says,
在第一个代码中，声明说，
int *B[10]
says that B is an array of 10 elements, each element of which is a pointer to a int. That is, B[i] is a int pointer and *B[i] is the integer it points to the first integer of the i-th saved text line.
说 B 是一个包含 10 个元素的数组，其中的每个元素都是一个指向 int 的指针。也就是说，B[i] 是一个 int 指针，*B[i] 是它指向第 i 个已保存文本行的第一个整数的整数。
回答by G_L
Using the really excellent example from Greggo, I got this to work as a bubble sort with passing an array as a pointer and doing a simple -1 manipulation.
使用来自 Greggo 的非常出色的示例，我将其用作冒泡排序，将数组作为指针传递并执行简单的 -1 操作。
#include<stdio.h>
void sub_one(int (*arr)[7])
{
int i;
for(i=0;i<7;i++)
{
(*arr)[i] -= 1 ; // subtract 1 from each point
printf("%i\n", (*arr)[i]);
}
}
int main()
{
int a[]= { 180, 185, 190, 175, 200, 180, 181};
int pos, j, i;
int n=7;
int temp;
for (pos =0; pos < 7; pos ++){
printf("\nPosition=%i Value=%i", pos, a[pos]);
}
for(i=1;i<=n-1;i++){
temp=a[i];
j=i-1;
while((temp<a[j])&&(j>=0)) // while selected # less than a[j] and not j isn't 0
{
a[j+1]=a[j]; //moves element forward
j=j-1;
}
a[j+1]=temp; //insert element in proper place
}
printf("\nSorted list is as follows:\n");
for(i=0;i<n;i++)
{
printf("%d\n",a[i]);
}
printf("\nmedian = %d\n", a[3]);
sub_one(&a);
return 0;
}
I need to read up on how to encapsulate pointers because that threw me off.
我需要阅读如何封装指针，因为这让我很失望。
回答by Jay Gupta
The argument of func is accepting double-pointer variable. Hope this helps...
func 的参数是接受双指针变量。希望这可以帮助...
#include <stdio.h>
int func(int **B){
}
int main(void){
int *B[10];
func(B);
return 0;
}
回答by Vaibhav Balkrishna Dhole
main()
{
int *arr[5];
int i=31, j=5, k=19, l=71, m;
arr[0]=&i;
arr[1]=&j;
arr[2]=&k;
arr[3]=&l;
arr[4]=&m;
for(m=0; m<=4; m++)
{
printf("%d",*(arr[m]));
}
return 0;
}
回答by utpal podder
In the function declaration you have to type as
在函数声明中，您必须键入
VOID FUN(INT *a[]);
/*HERE YOU CAN TAKE ANY FUNCTION RETURN TYPE HERE I CAN TAKE VOID AS THE FUNCTION RETURN TYPE FOR THE FUNCTION FUN*/
//IN THE FUNCTION HEADER WE CAN WRITE AS FOLLOWS
void fun(int *a[])
//in the function body we can use as
a[i]=var
