将int值返回到另一个funciton

发布于 2025-02-08 19:41:41 字数 453 浏览 1 评论 0原文

我是C的完整初学者,很抱歉提出一个琐碎的问题。 问:我想将在number_generator函数中计算的Random_num变量的值返回到主函数:

int main() //didn't paste headers and prototype 
{
   int random_num; // <= to here
   number_generator(random_num);
   printf("random number: %d", random_num);

   return 0;
}

int number_generator(int random_num)
{
   srand(time (0));
   random_num = (rand() % 100) + 0; // <= return value from here
} 

当前,printf的终端输出为0,而应在0到100范围内。

I am a complete beginner in C and I'm sorry in advance for a trivial question.
Q: I would like to return the value of random_num variable that is calculated in number_generator function to the main function:

int main() //didn't paste headers and prototype 
{
   int random_num; // <= to here
   number_generator(random_num);
   printf("random number: %d", random_num);

   return 0;
}

int number_generator(int random_num)
{
   srand(time (0));
   random_num = (rand() % 100) + 0; // <= return value from here
} 

Currently, the terminal output of printf is 0 while it should be in a range from 0 to 100.

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赤濁 2025-02-15 19:41:41

这样做以下方法

int number_generator(void )

int main() //didn't paste headers and prototype 
{
   int random_num = number_generator(); // <= to here
   printf("random number: %d", random_num);

   return 0;
}

int number_generator(void )
{
   srand(time (0));
   return (rand() % 100) + 0; // <= return value from here
}

是另一种方法是

void number_generator(int *random_num);

int main() //didn't paste headers and prototype 
{
   int random_num; // <= to here
   number_generator( &random_num);
   printf("random number: %d", random_num);

   return 0;
}

void number_generator(int *random_num)
{
   srand(time (0));
   *random_num = (rand() % 100) + 0; // <= return value from here
} 

要注意在此表达式中添加0的方法

(rand() % 100) + 0

是没有意义的。相当于

(rand() % 100)

Do it the following way

int number_generator(void )

int main() //didn't paste headers and prototype 
{
   int random_num = number_generator(); // <= to here
   printf("random number: %d", random_num);

   return 0;
}

int number_generator(void )
{
   srand(time (0));
   return (rand() % 100) + 0; // <= return value from here
}

Another approach is the following

void number_generator(int *random_num);

int main() //didn't paste headers and prototype 
{
   int random_num; // <= to here
   number_generator( &random_num);
   printf("random number: %d", random_num);

   return 0;
}

void number_generator(int *random_num)
{
   srand(time (0));
   *random_num = (rand() % 100) + 0; // <= return value from here
} 

Pay attention to that adding 0 in this expression

(rand() % 100) + 0

does not make a sense. It is equivalent to

(rand() % 100)
~没有更多了~
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