简单的C随机彩票号码

Question

我想以最简单的方式生成 5 个不同的数字并将它们放入数组中。我的思维方式有问题，你能纠正我的代码吗？

void lottery(int *array){
int i = 0;
while(i != 5){
bool good = true;
int number = rand()%90+1;
for(int j=0; j<5; j++){
if(array[j] == number)
good = false;
break;
}
if(good){
array[i] == number;
i = i+1;
}
}
}
int main(){
srand(time(0));
int numbers[5];
lottery(numbers);
for(int i =0; i<5; i++){
printf("%d, ",numbers[i]);
}
return 0;
}

Answer 1

包括kingW3和rici的发现，并稍微清理一下：

// compiles without errors with:
// clang -O3 -g3 -Weverything -W -Wall -Wextra -Wpedantic -fsanitize=bounds  -std=c11  -o stacktest stacktest.c 
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <time.h>
void lottery(int *array);
void lottery(int *array){
int i = 0;
while(i != 5){
bool good = true;
int number = rand()%90+1;
// limit is 'i' because you fill the array in order
for(int j=0; j<i; j++){
// Would always `break` without the curly brackets
if(array[j] == number){
good = false;
break;
}
}
if(good){
// Use '=' instead of '=='
array[i] = number;
i = i+1;
}
}
}
int main(void){
// time returns type 'type_t', needs casting to get rid of
// warning
srand( (unsigned int) time(0));
// return of rand() is always >= 0, so 
// initialize with a negative number
int numbers[5] = {-1};
int i;
lottery(numbers);
for(i =0; i<4; i++){
printf("%d, ",numbers[i]);
}
printf("%dn",numbers[i]);
return 0;
}

缺少括号是另一个错误。

警告：time()的分辨率很可能为一秒，所以不要按顺序运行得太快，否则你会得到相同的数字。