C语言 初始化使指针从整数而不进行强制转换 - C
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/33404478/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Initialization makes pointer from integer without a cast - C
提问by Jeff H
Sorry if this post comes off as ignorant, but I'm still very new to C, so I don't have a great understanding of it. Right now I'm trying to figure out pointers.
对不起，如果这篇文章被认为是无知的，但我对 C 仍然很陌生，所以我对它没有很好的理解。现在我正试图找出指针。
I made this bit of code to test if I can change the value of b in the change function, and have that carry over back into the main function(without returning) by passing in the pointer.
我编写了这段代码来测试我是否可以在更改函数中更改 b 的值，并通过传入指针将其带回到主函数中（不返回）。
However, I get an error that says.
但是，我收到一条错误消息。
Initialization makes pointer from integer without a cast
int *b = 6
From what I understand,
据我了解，
#include <stdio.h>
int change(int * b){
* b = 4;
return 0;
}
int main(){
int * b = 6;
change(b);
printf("%d", b);
return 0;
}
Ill I'm really worried about is fixing this error, but if my understanding of pointers is completely wrong, I wouldn't be opposed to criticism.
我真的很担心修复这个错误，但如果我对指针的理解完全错误，我不会反对批评。
回答by Soren Goyal
To make it work rewrite the code as follows -
要使其工作，请按如下方式重写代码 -
#include <stdio.h>
int change(int * b){
* b = 4;
return 0;
}
int main(){
int b = 6; //variable type of b is 'int' not 'int *'
change(&b);//Instead of b the address of b is passed
printf("%d", b);
return 0;
}
The code above will work.
上面的代码将起作用。
In C, when you wish to change the value of a variable in a function, you "pass the Variable into the function by Reference". You can read more about this here - Pass by Reference
在 C 中，当您希望更改函数中变量的值时，您可以“通过引用将变量传递给函数”。您可以在此处阅读更多相关信息 -通过引用传递
Now the error means that you are trying to store an integer into a variable that is a pointer, without typecasting. You can make this error go away by changing that line as follows (But the program won't work because the logic will still be wrong )
现在错误意味着您正在尝试将一个整数存储到一个指针变量中，而不进行类型转换。您可以通过如下更改该行来消除此错误（但程序将无法运行，因为逻辑仍然是错误的）
int * b = (int *)6; //This is typecasting int into type (int *)
回答by BobRun
Maybe you wanted to do this:
也许你想这样做：
#include <stdio.h>
int change( int *b )
{
*b = 4;
return 0;
}
int main( void )
{
int *b;
int myint = 6;
b = &myint;
change( &b );
printf( "%d", b );
return 0;
}
回答by milevyo
#include <stdio.h>
int change(int * b){
* b = 4;
return 0;
}
int main(){
int b = 6; // <- just int not a pointer to int
change(&b); // address of the int
printf("%d", b);
return 0;
}
回答by udarH3
Maybe too late, but as a complement to the rest of the answers, just my 2 cents:
也许为时已晚，但作为对其余答案的补充，只需我的 2 美分：
void change(int *b, int c)
{
*b = c;
}
int main()
{
int a = 25;
change(&a, 20); --> with an added parameter
printf("%d", a);
return 0;
}
In pointer declarations, you should only assign the address of other variables e.g "&a"..
在指针声明中，您应该只分配其他变量的地址，例如“&a”..
