如何生成n个随机1S在C / C ++的无符号的字符数组？ [英] How to generate n random 1s in an unsigned char array in c/c++?
问题描述
这是一个我刚才问不同的问题，这是更具挑战性。
我有一个无符号的字符数组，说
unsigned char型A [16]。
我需要生成，我将适用于我的数组A [16]面具载体。
它应该包含的'1'，在0℃n个; N'LT; 16 * 8（掩模载体可以是一个排列B [16]只要有n阵列中的'1'编号）
我还需要在向量随机分布的1的这些n个。
我怎样才能做到这一点在C / C ++？
感谢您！
编辑：
我的想法如下：
我将产生n个（做检查需求，以确保所有的n号不相同）的随机数，并将其存储在数组TMP [n]的。然后面膜是基于移动产生的。
函数srand（时间（0））;
对于（i = 0; I＆LT; N;我++）{
为（J = 0; J＆LT; I; J ++）
而（TMP [I] == TMP [J]）//以确保所有N个随机数是不同的
TMP [I] =兰特（）％128;unsigned char型面膜[16]
对于（i = 0; I＆LT; N;我++）
掩模[16] | =（1 <<;＆下; TMP [I]）; //生成面具
生成随机（I，J）对数字，其中的I＆LT ; 16 和 J＆LT; 8 。如果在位置的位 B [I]及（1 LT;＆LT; j）条未设置，设置和增加计数。循环直到算变为N。
code的位（未经测试）：
无效generate_n_bit_mask（unsigned char型B〔]，INT N）
{
//避免以后无限循环。
的for（int i = 0;（I＆LT; 16）; ++ I）{
B〔Ⅰ〕= 0;
}
//不变：k是当前掩蔽的位数。
对于（INT K = 0;（K＆LT; N）;）
{
//随机选择位。
INT I =兰特（）％16;
INT J =兰特（）％8;
unsigned char型面膜= 1＆LT;＆LT;焦耳;
//如果不选择previously设置。
如果（（B [1] - 安培;面罩）== 0）{
B〔I] | =面具，++ K表;
}
}
}
锻炼，接受挑战：去掉魔法恒 16 从code
修改：本修饰您的意见建议包含了一个讨厌的错误。下面是一个测试程序与位分布在输出面具的方式打球。
的#include＆LT;＆iostream的GT;
＃包括LT＆;＆了iomanip GT;
＃包括LT＆;＆的ctime GT;无效generate_n_bit_mask（unsigned char型B〔]，INT N）
{
//避免以后无限循环。
的for（int i = 0;（I＆LT; 16）; ++ I）{
B〔Ⅰ〕= 0;
}
//不变：k是当前掩蔽的位数。
对于（INT K = 0;（K＆LT; N）;）
{
//随机选择位。
INT I =的std ::兰特（）％16;
INT J =的std ::兰特（）％8;
unsigned char型面膜= 1＆LT;＆LT;焦耳;
//如果不选择previously设置。
如果（（B [1] - 安培;面罩）== 0）{
B〔I] | =面具，++ K表;
}
}
INT J = 0;
}//一个字节中设置位的计数。
INT BIT_COUNT（unsigned char型X）
{
INT N = 0;
的for（int i = 0;（I＆LT; 8）; ++ I）{
的n + =（（X＆GT; I标记）及1）;
}
返回（N）;
}//在16字节比特集合计数。
INT total_bit_count（unsigned char型B〔]）
{
INT N = 0;
的for（int i = 0;（I＆LT; 16）; ++ I）{
N + = BIT_COUNT（B [I]）;
}
返回（N）;
}INT主（INT，CHAR **）
{
的std ::函数srand（的std ::时间（0））;
unsigned char型B〔16〕;
//对于n的所有可能的值
的for（int i = 0;（I＆LT; = 16 * 8）; ++ I）
{
//生成带N位设置一个16字节的面具。
generate_n_bit_mask（B，I）;
//验证n位被设置。
INT N = total_bit_count（B）;
如果（N！= 1）{
性病::法院LT＆;＆LT; I＆LT;＆LT; ：＆所述;＆下; N'LT;＆LT;的std :: ENDL;
}
}
}
在运行此程序，它会尝试 N 从 0 到 N 位，然后验证准确 N 位设置。如果出现任何错误（为 N ，有些 K！= N 位设置一定的价值），一条消息输出。
如果我更改了条件 IF（（B [I] ^面膜）！= 0），我得到的输出一致的错误。每次运行产生至少1的错误消息。原状 IF（（B [1] - 安培;面罩）== 0）。一贯产生0的错误消息
This is a different question from the one I just asked and it is more challenging.
I have an unsigned char array, say unsigned char A[16]. I need to generate a mask vector which i will apply to my array A[16].
It should contain n number of '1's, where 0 < n < 16*8 (The mask vector can be an array B[16] as long as there are n number of '1's in the array)
I also need these n number of '1's distributed randomly in the vector.
How can I do this in c/c++?
Thank you!
Edit: My thought is as follows: I will generate n random numbers (checking needs to be done to make sure all n numbers are not the same) and store them in array tmp[n]. Then mask is generated based on shifting.
srand(time(0));
for(i = 0; i < n; i++){
for(j = 0; j < i; j++)
while(tmp[i] == tmp[j]) // to make sure all n random numbers are different
tmp[i] = rand()%128;
unsigned char mask[16]
for(i = 0; i < n; i++)
mask[16] |= (1 << tmp[i]); //generate mask
Generate random (i,j) pair of numbers, where i < 16 and j < 8. If the bit at position B[i]&(1<<j) is not set, set it and increment "count". Loop until "count" reaches "n".
A bit of code (untested):
void generate_n_bit_mask ( unsigned char B[], int n )
{
// avoid infinite loop later on.
for ( int i=0; (i < 16); ++i ) {
B[i] = 0;
}
// invariant: k is number of currently masked bits.
for ( int k = 0; (k < n); )
{
// select bit at random.
int i = rand() % 16;
int j = rand() % 8;
unsigned char mask = 1 << j;
// set it if not selected previously.
if ( (B[i]&mask) == 0 ) {
B[i] |= mask, ++k;
}
}
}
Exercise, for the challenge: remove magic constant 16 from the code.
Edit: The modification suggested in your comments contains a nasty bug. Here is a test program to play with the way bits are distributed in your output mask.
#include <iostream>
#include <iomanip>
#include <ctime>
void generate_n_bit_mask ( unsigned char B[], int n )
{
// avoid infinite loop later on.
for ( int i=0; (i < 16); ++i ) {
B[i] = 0;
}
// invariant: k is number of currently masked bits.
for ( int k = 0; (k < n); )
{
// select bit at random.
int i = std::rand() % 16;
int j = std::rand() % 8;
unsigned char mask = 1 << j;
// set it if not selected previously.
if ( (B[i]&mask) == 0 ) {
B[i] |= mask, ++k;
}
}
int j = 0;
}
// count number of set bits in a byte.
int bit_count ( unsigned char x )
{
int n = 0;
for ( int i = 0; (i < 8); ++i ) {
n += ((x >> i) & 1);
}
return (n);
}
// count number of set bits in 16 bytes.
int total_bit_count ( unsigned char B[] )
{
int n = 0;
for ( int i = 0; (i < 16); ++i ) {
n += bit_count(B[i]);
}
return (n);
}
int main ( int, char ** )
{
std::srand(std::time(0));
unsigned char B[16];
// for all possible values of "n"
for ( int i = 0; (i <= 16*8); ++i )
{
// generate a 16 byte mask with "n" set bits.
generate_n_bit_mask(B, i);
// verify that "n" bits are set.
int n = total_bit_count(B);
if ( n != i ) {
std::cout << i << ": " << n << std::endl;
}
}
}
When this program is run, it will try every value of n from 0 to 16*8 and generate a random mask with n bits, then verify that exactly n bits are set. If any error occurs (for some value of n, some k!=n bits are set), a message is output.
If I change the condition to if ( (B[i]^mask) != 0 ), I get consistent errors in the output. Every run produces at least 1 error message. The original condition if ( (B[i]&mask) == 0 ) consistently produces 0 error messages.
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