这是个随机产生一个'A'~'Z'的字符串的程序~注释部分的91和65应怎样改啊~?我有搞不懂ASCII码的转换`

#include<iostream>
#include<ctime>
using namespace std;

void main()
{
int i,j;
char *charp;
srand(time(0));
i=rand()%10+1;
charp=new char[i];
for(j=0;j<i;j++)
{

charp[j]=rand()%91+65; //这里应该怎么改啊?
cout<<charp[j];
}
cout<<"\n";
}

if rand()>65 it is ok
if rand()<65 you can make it >65 then every thing is ok

什么~??~不懂你的意思~~~

change
charp[j]=rand()%91+65;
to
charp[j] = rand()%26 + 'A';
这样就算明天VC决定ASCII的字符表变了，'A'不再是65了你的程序也不需要重写。

[此贴子已经被作者于2006-4-10 0:38:57编辑过]

那么如果大小写字母混合产生~应该怎么改呢~

change
charp[j] = rand()%26 + 'A';
to
charp[j] = rand()%52+ 'A';

多加点长度就行了

简单的改为52,有问题:
因为A~Z的ASCII码为65-90
a-z的ASCII码为97-122
只要再处理一下就行了

int intRrand = rand() % 52;

if (intRand < 26)
charp[j] = intRand + 'A';
else
charp[j] = intRand - 26 + 'a';

不行啊~everajax~~
只产生p一个字符啊

#include "stdafx.h"
#include<iostream>
#include<ctime>
using namespace std;

int main(int argc, char* argv[])
{
int i,j;
char *charp;
srand(time(0));
i=rand()%10+1;
charp=new char[i];
for(j=0;j<i;j++)
{
int intRand = rand() % 52;

if (intRand < 26)
charp[j] = intRand + 'A';
else
charp[j] = intRand - 26 + 'a';

cout<<charp[j];
}
cout<<"\n";
return 0;
}


aerydJ
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