警告：格式为“％d”，预期类型为“ 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

因此，我是C语言的新手，并且对于此警告的内容遇到了麻烦。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\n");
  scanf("%s\n", name);
  printf("input your age\n");
  scanf("%d\n", age);
  printf("%s %d\n", name, age);

}

推荐答案

code> scanf 函数将变量的地址放入结果中。

编写 scanf（ ％d， & someVar）将通过 someVar <的地址 / code>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)

当您编写 scanf（％d，age）时，您会通过 age 变量的 value 到 scanf 。它将尝试将一个数字放入地址 0 的内存中（因为 age 是 0 ），并弄得一团糟。

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.

您需要通过& age 到 scanf 。

您还需要为 scanf 将字符串读入名称：

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

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

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