C语言中的指针和数组之间的区别（） _sizeof

本文概述

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指针用于存储动态分配的数组的地址, 以及用于作为参数传递给函数的数组。在其他情况下, 数组和指针是两个不同的东西, 请参见以下程序以证明此语句的合理性。
sizeof运算符的行为
C

// 1st program to show that array and pointers are different #include < stdio.h> int main() { int arr[] = { 10, 20, 30, 40, 50, 60 }; int * ptr = arr; // sizof(int) * (number of element in arr[]) is printed printf ( "Size of arr[] %ld\n" , sizeof (arr)); // sizeof a pointer is printed which is same for all // type of pointers (char *, void *, etc) printf ( "Size of ptr %ld" , sizeof (ptr)); return 0; }

C ++

// 1st program to show that array and pointers are different #include < iostream> using namespace std; int main() { int arr[] = { 10, 20, 30, 40, 50, 60 }; int * ptr = arr; // sizof(int) * (number of element in arr[]) is printed cout < < "Size of arr[] " < < sizeof (arr) < < "\n" ; // sizeof a pointer is printed which is same for all // type of pointers (char *, void *, etc) cout < < "Size of ptr " < < sizeof (ptr); return 0; }

输出如下

Size of arr[] 24 Size of ptr 8

不允许将任何地址分配给数组变量。
C

// IInd program to show that array and pointers are different #include < stdio.h> int main() { int arr[] = {10, 20}, x = 10; int *ptr = & x; // This is fine arr = & x; // Compiler Error return 0; }

输出如下：

Compiler Error: incompatible types when assigning to type 'int[2]' from type 'int *'

【C语言中的指针和数组之间的区别（）】见
以前的帖子
关于此主题的更多差异。
尽管数组和指针是不同的东西, 但是数组的以下属性使它们看起来相似。

数组名称给出数组第一个元素的地址。

例如, 考虑以下程序。
C

#include < stdio.h> int main() { int arr[] = { 10, 20, 30, 40, 50, 60 }; // Assigns address of array to ptr int * ptr = arr; printf ( "Value of first element is %d" , *ptr); return 0; }

C ++

// 1st program to show that array and pointers are different #include < iostream> using namespace std; int main() { int arr[] = { 10, 20, 30, 40, 50, 60 }; // Assigns address of array to ptr int * ptr = arr; cout < < "Value of first element is " < < *ptr; return 0; }

输出如下

Value of first element is 10

使用指针算法访问数组成员。
编译器使用指针算法访问数组元素。例如, 像" arr [i]"之类的表达式被编译器视为*(arr + i)。这就是为什么像*(arr + i)这样的表达式可用于数组arr, 而像ptr [i]这样的表达式也可用于指针ptr的原因。
C

#include < stdio.h> int main() { int arr[] = {10, 20, 30, 40, 50, 60}; int *ptr = arr; printf ( "arr[2] = %d\n" , arr[2]); printf ( "*(arr + 2) = %d\n" , *(arr + 2)); printf ( "ptr[2] = %d\n" , ptr[2]); printf ( "*(ptr + 2) = %d\n" , *(ptr + 2)); return 0; }

C ++

#include < iostream> using namespace std; int main() { int arr[] = { 10, 20, 30, 40, 50, 60 }; int * ptr = arr; cout < < "arr[2] = " < < arr[2] < < "\n" ; cout < < "*(arr + 2) = " < < *(arr + 2) < < "\n" ; cout < < "ptr[2] = " < < ptr[2] < < "\n" ; cout < < "*(ptr + 2) = " < < *(ptr + 2) < < "\n" ; return 0; }

输出如下

arr[2] = 30 *(arr + 2) = 30 ptr[2] = 30 *(ptr + 2) = 30

数组参数始终作为指针传递, 即使我们使用方括号也是如此。
C

#include < stdio.h> int fun( int ptr[]) { int x = 10; // size of a pointer is printed printf ( "sizeof(ptr) = %d\n" , ( int ) sizeof (*ptr)); // This allowed because ptr is a pointer, not array ptr = & x; printf ( "*ptr = %d " , *ptr); return 0; } // Driver code int main() { int arr[] = { 10, 20, 30, 40, 50, 60 }; // size of a array is printed printf ( "sizeof(arr) = %d\n" , ( int ) sizeof (arr)); fun(arr); return 0; }

输出如下