警告:格式“%d"需要类型“int *"，但参数 2 的类型为“int" [英] warning: format ‘%d’ expects type ‘int *’, but argument 2 has type ‘int’

问题描述

所以我是 C 的新手，并且对这个警告发生的事情有疑问.警告是什么意思，我该如何解决.我写的代码在这里:

So I'm new to C and am having trouble with whats happening with this warning. What does the warning mean and how can i fix it. The code i wrote is here:

void main(void)
{
  char* name = "";
  int age = 0;
  printf("input your name
");
  scanf("%s
", name);
  printf("input your age
");
  scanf("%d
", age);
  printf("%s %d
", name, age);

}

推荐答案

scanf 函数将变量的地址 放入其中.
编写 scanf("%d", &someVar) 将传递 someVar 的 address 变量(使用 & 一元运算符).
scanf 函数会将一个数字放入该地址的内存中.(其中包含您的变量)

The scanf function takes the address of a variable to put the result into.
Writing scanf("%d", &someVar) will pass the address of the someVar variable (using the & unary operator).
The scanf function will drop a number into the piece of memory at that address. (which contains your variable)

当你写scanf("%d", age)时，你将age变量的值传递给scanf.它会尝试将一个数字放入地址 0 的内存块中(因为 age 是 0)，然后一团糟.

When you write scanf("%d", age), you pass the value of the age variable to scanf. It will try to drop a number into the piece of memory at address 0 (since age is 0), and get horribly messed up.

您需要将 &age 传递给 scanf.

You need to pass &age to scanf.

您还需要为 scanf 分配内存以将字符串读入 name:

You also need to allocate memory for scanf to read a string into name:

char name[100];
scanf("%99s
", name);

这篇关于警告:格式“%d"需要类型“int *"，但参数 2 的类型为“int"的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！