字符串生成 [英] string generation
问题描述
我想以编程方式生成每个可能的16个字符组合的
字符0-9，A-Z和a-z。我目前的代码如下：
#include< stdio.h>
#include< ctype.h>
int main（）{
char strBuf [17] = {0};
int i，j;
for（i = 0; i< 16; i ++）{
for（j =''0''; j< =''z''; j ++）{
如果（！isalnum（j））继续;
strBuf [i] = j;
printf（"％s \ n"，strBuf） ;
}
}
返回0;
}
只有最后一个字符递增。我理解我的问题，我可以用b $ b来修复数组中每个空格的for循环，但是我想知道
如果有更好的话解决方案。
Ken Human< ke ****** @ comcast.net>写道：
我想以编程方式生成
字符0-9，A-Z和a-z的每个可能的16个字符组合。我现在的代码如下：
真的，每一个？
你希望在哪台电脑上运行这个？
< br $> b $ b -
Chris。
Ken Human写道：我想要生成每个可能的16个字符组合的
字符0-9，AZ和az以编程方式。
你确定吗？那是'pow（62,16），这是很多组合。
关于4.8e28，我想。
我现在的代码如下：
#include< stdio.h>
#include< ctype.h>
int main（）{
char strBuf [17] = {0} ;对于（i = 0; i< 16; i ++）{
for（j =''0''; j< = ''z''; j ++）{
如果（！isalnum（j））继续;
strBuf [i] = j;
printf（"％s \ n"，strBuf） ）;
}
返回0;
}
只有最后一个字符递增。我理解我的问题，我可以用数组中每个空格的for循环修复它，但我想知道
是否有更好的解决方案。
未经测试的代码块：
{
char foo [] =" 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
" abcdefghijklmnopqrstuvwxyz" ;;
unsigned nfoos = sizeof foo / sizeof * foo;
unsigned ndx [16] = {0}，i;
while （ndx [15]！= nfoos）
{
for（i = 0; i< 16; i ++）putchar（foo [ndx [i]]）;
putchar（''\ n''）;
for（i = 0; i< 16; i ++）
{
ndx [i] ++;
如果（ndx [i]< nfoos）中断;
ndx [i] = 0; < br $>
}
}
}
Chris McDonald写道：Ken Human< ke ****** @ comcast.net>写道：
我想以编程方式生成
字符0-9，A-Z和a-z的每个可能的16个字符组合。我现在的代码如下：
真的，每一个？
你希望在哪台电脑上运行这个？
我从来都不擅长数学，是可能组合的数量
62 ^ 16？谢谢你的关心。我在Plasmo Mag-8上运行它，
可以在1秒内计算出这个确切的数据量。不用说
说，它也有无限的内存。
让我们说我想要一个4个字符的组合。
I want to generate every possible 16 character combination of the
characters 0-9, A-Z, and a-z programatically. My current code follows:
#include <stdio.h>
#include <ctype.h>
int main() {
char strBuf[17] = {0};
int i, j;
for(i = 0; i < 16; i++) {
for(j = ''0''; j <= ''z''; j++) {
if(!isalnum(j)) continue;
strBuf[i] = j;
printf("%s\n", strBuf);
}
}
return 0;
}
Only the last character increments. I understand my problem and I can
fix it with a for loop for each space in the array, but I''d like to know
if there''s a better solution.
Ken Human <ke******@comcast.net> writes:
I want to generate every possible 16 character combination of the
characters 0-9, A-Z, and a-z programatically. My current code follows:
Really, every one?
And on what computer are you hoping to run this?
--
Chris.
Ken Human wrote:I want to generate every possible 16 character combination of the
characters 0-9, A-Z, and a-z programatically.
Are you sure? That''s pow(62,16), which is a lot of combinations.
About 4.8e28, I think.
My current code follows:
#include <stdio.h>
#include <ctype.h>
int main() {
char strBuf[17] = {0};
int i, j;
for(i = 0; i < 16; i++) {
for(j = ''0''; j <= ''z''; j++) {
if(!isalnum(j)) continue;
strBuf[i] = j;
printf("%s\n", strBuf);
}
}
return 0;
}
Only the last character increments. I understand my problem and I can
fix it with a for loop for each space in the array, but I''d like to know
if there''s a better solution.
untested code block:
{
char foo[]="0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
"abcdefghijklmnopqrstuvwxyz";
unsigned nfoos = sizeof foo/sizeof *foo;
unsigned ndx[16] = {0}, i;
while(ndx[15] != nfoos)
{
for (i = 0; i < 16; i++) putchar(foo[ndx[i]]);
putchar(''\n'');
for(i = 0; i < 16; i++)
{
ndx[i]++;
if (ndx[i] < nfoos) break;
ndx[i] = 0;
}
}
}
Chris McDonald wrote:Ken Human <ke******@comcast.net> writes:I want to generate every possible 16 character combination of the
characters 0-9, A-Z, and a-z programatically. My current code follows:
Really, every one?
And on what computer are you hoping to run this?
I was never very good at math, is the number of possible combinations
62^16? Thank you for your concern. I''m running it on the Plasmo Mag-8,
which can calculate this exact amount of data in 1 second. Needless to
say, it also has an unlimited amount of memory.
Let''s say that I want a 4 character combination.
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