【发布时间】:2015-05-25 23:38:00
【问题描述】:
我试图了解 malloc 和字符数组(c 样式)是如何工作的。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,
- 为什么 charBuffer 的大小是 8(见输出的第一行),虽然我分配了 10 的大小?
- 为什么我可以向 charBuffer 添加 15 个字符,尽管我使用 malloc 只为 10 个字符分配了内存? (参见代码第 4 节)
- 为什么在第 5 节的输出中打印的是参考索引之后的字符而不是相应字符的地址?
- 如何找到单个字符的地址?
- 当字符数组的元素被填满时,是否可以知道数组的大小?例如在第 3 节的循环中显示 sizeof(charbuffer),我们应该得到 1,2,3..,10?
【问题讨论】:
-
天啊...每个问题一个问题,拜托!!
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@LightnessRacesinOrbit 很抱歉。但是,当上下文相同时,可以发布 5 个背靠背问题吗?
标签: c++ arrays pointers char malloc