C语言 如何使用malloc在ANSI-C中声明动态整数数组并将输入整数放入其中？
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How to declare a dynamic integer array in ANSI - C using malloc and put input integers into it?
提问by TomG
First I want the user to input what is the size of the desired array. So I am using:
首先，我希望用户输入所需数组的大小。所以我正在使用：
int size;
scanf("&d",&size);
Now I want to create an integer array using a pointer and the malloc function. This is what I did:
现在我想使用指针和 malloc 函数创建一个整数数组。这就是我所做的：
int *p1 = (int*)malloc(sizeof(int)*size);
According to my understanding, this is like using:
根据我的理解，这就像使用：
int p1[size];
But how do I use it like an array?
但是我如何像数组一样使用它？
Question1: Now I want the user to input as many integers as he wrote into this "array". But I can't use p[0] because it is not an array, it is a pointer.
问题1：现在我希望用户输入与他写入这个“数组”一样多的整数。但是我不能使用 p[0] 因为它不是一个数组，它是一个指针。
Question2: I want to "send" this array to a function that gets an array of integers. So again, this is not an array, how can I "give" it to the function?
问题2：我想将此数组“发送”给一个获取整数数组的函数。再说一次，这不是一个数组，我怎么能把它“给”给函数呢？
回答by A human being
Answer to first question:
回答第一个问题：
for(i = 0; i < size; i++ )
{
scanf("%d",&p[i]);
/*p[i] is the content of element at index i and &p[i] is the address of element
at index i */
}
Or
或者
for(i = 0; i < size; i++ )
{
scanf("%d",(p+i)); //here p+i is the address of element at index i
}
Answer to second question:
回答第二个问题：
For sending this array to the function, just call the function like this:
要将这个数组发送到函数，只需像这样调用函数：
function(p); //this is sending the address of first index of p
void function( int *p ) //prototype of the function
回答by Lefteris E
- Question 1: 1d arrays and pointers to properly allocated memory are pretty-much the same thing.
- Question 2: When passing an array to a method you are actually passing the address of the 1st element of that array
- 问题 1：一维数组和指向正确分配内存的指针几乎是一回事。
- 问题 2：将数组传递给方法时，您实际上是在传递该数组的第一个元素的地址
An array is actually a pointer to the first element of the array
数组实际上是指向数组第一个元素的指针
回答by bash.d
You can use the subscript-syntax to access an element of your pointer.
您可以使用subscript-syntax 来访问指针的元素。
p1[3] = 5; // assign 5 to the 4th element
But, this syntax is actually converted into the following
但是，这个语法实际上被转换成以下
*(p1+3) = 5; // pointer-syntax
For your second question, define a function and pass a pointer
对于你的第二个问题，定义一个函数并传递一个指针
int* getarray(int* p1, size_t arraysize){ } //define the function
int* values = getarray(p1, size); // use the function
回答by Pulkit Agarwal
sorry for bothering everyone but Miss Upasana was right this is the correct method for dynamic array use. After declaring ur array through malloc u can directly use it exactly as array as follows::
抱歉打扰了大家，但 Upasana 小姐是对的，这是动态数组使用的正确方法。通过 malloc 声明你的数组后，你可以直接像数组一样使用它，如下所示：
for(int i = 0; i < size; i++ )
{
scanf("%d",p+i);
/*p+i denoting address of memory allocated by malloc */
}
Second Answer: Now simply pass this address to any function use address to find values like:
第二个答案：现在只需将此地址传递给任何函数使用地址来查找值，例如：
function(int *p)
/* access as &p for first value &p+2 for second p+4 for third and so on*/
