我试图理解malloc和字符数组(c样式)是如何工作的。请考虑以下代码,
// Example program
#include <iostream>
#include <cstdlib>
#include <iomanip>
using namespace std;
int main()
{
//section1: Init
char line0[10] = {'a','b','c','d','e','f','g','h','i','j'};
char line1[10] = {'z','y','x','w','v','u','t','s','r','q'};
//section2: Allocate Character array
char* charBuffer = (char*) malloc(sizeof(char)*10);
cout<<sizeof(charBuffer)<<endl;
//Section3: add characters to the array
for (int i=0;i<10;i++)
{
*&charBuffer[i] = line0[i];
}
//Section4:-add character to array using pointers
for (int i=0;i<15;i++)
{
charBuffer[i] = line1[i%10];
}
//section5:-address of characters in the array
for (int i=0;i<15;i++)
{
cout<<"Address of Character "<<i<<" is: "<<&charBuffer[i]<<"\n";
}
char *p1;
p1 = &charBuffer[1];
cout<<*p1<<endl;
cout<<charBuffer<<endl;
free(charBuffer);
return 0;
}
输出: -
8
Address of Character 0 is: zyxwvutsrqzyxwv
Address of Character 1 is: yxwvutsrqzyxwv
Address of Character 2 is: xwvutsrqzyxwv
Address of Character 3 is: wvutsrqzyxwv
Address of Character 4 is: vutsrqzyxwv
Address of Character 5 is: utsrqzyxwv
Address of Character 6 is: tsrqzyxwv
Address of Character 7 is: srqzyxwv
Address of Character 8 is: rqzyxwv
Address of Character 9 is: qzyxwv
Address of Character 10 is: zyxwv
Address of Character 11 is: yxwv
Address of Character 12 is: xwv
Address of Character 13 is: wv
Address of Character 14 is: v
y
zyxwvutsrqzyxwv
我想了解以下内容,
答案 0 :(得分:4)
为什么charBuffer 8的大小(见第一行输出)虽然我已经分配了10的大小?
不是。你打印出指向该缓冲区的指针的大小。
为什么我能够为charBuffer添加15个字符虽然我已经使用malloc为10个字符分配了内存?
你不是。但是,相反,允许计算机不要忘记告知您的错误。你违反了记忆规则。
为什么打印参考索引后的字符而不是第5部分输出中相应字符的地址?
因为将char*
插入流会触发格式化插入,因此流假定您正在流式传输C字符串。哪个,你 。
如何找到单个字符的地址?
您可以编写static_cast<void*>(&charBuffer[i])
以避免此特殊情况处理,并改为打印出地址。
当字符数组的元素被填充时,可以知道数组的大小吗?例如,在第3节的循环中显示sizeof(charbuffer),我们应该得到1,2,3 ..,10?
数组的大小永远不会改变,只会改变您为其写入新值的元素数。您可以使用计数器变量自己跟踪。