ç冲突的类型错误 [英] C conflicting type bug
问题描述
当我尝试编译此我被GCC,我有冲突的类型 pack_cui(C)告诉
。
我不明白我怎么可以让冲突的类型,因为我知道我传递一个的char *
。我是新的C,所以我敢肯定,我失去了一些东西明显。
INT的main(){
字符* C =的malloc(sizeof的(字符)* 4);
C [0] = 1;
C [1] = 2;
C [2] = 1;
C [3] = 4;
pack_cui(℃);
返回0;
}unsigned int类型pack_cui(字符* C){
unsigned int类型new_int = 0;
无符号整型我;
对于(I = 0; I&下; 4;我++){
new_int = new_int | (无符号整数)(符号int)C [I]
如果(ⅰ= 3!)new_int = new_int&所述;&下; 8;
}
返回new_int;
}
我收到的错误是
hw12.c:17:14:错误:冲突的类型'pack_cui
unsigned int类型pack_cui(字符* C){
^
hw12.c:13:5:注意:previous的pack_cui'隐式声明在这里
pack_cui(℃);
当你调用一个未声明的功能,C89标准要求隐式声明。该声明将是:
INT pack_cui();
的()
是不一样的(无效),该函数接受一个
,表示未指定的若干参数,而(无效)
表示为零。这是从pre-ISO / ANSI C遗留下来的,早在K&放大器; R天。
您不希望出现这种情况,因为这是错误的声明。在上面创建自己的声明,上面的main()
:
无符号pack_cui(字符*);
When I try to compile this I'm being told by gcc that I have conflicting types for pack_cui(C)
.
I don't see how I could be getting conflicting types as I know I'm passing in a char*
. I'm new to C so I'm sure I'm missing something obvious.
int main(){
char* C = malloc(sizeof(char) * 4);
C[0] = 1;
C[1] = 2;
C[2] = -1;
C[3] = 4;
pack_cui(C);
return 0;
}
unsigned int pack_cui(char* C){
unsigned int new_int = 0;
unsigned int i;
for(i = 0; i < 4; i++){
new_int = new_int | (unsigned int)(signed int)C[i];
if(i != 3) new_int = new_int << 8;
}
return new_int;
}
The error I received was
hw12.c:17:14: error: conflicting types for ‘pack_cui’
unsigned int pack_cui(char* C){
^
hw12.c:13:5: note: previous implicit declaration of ‘pack_cui’ was here
pack_cui(C);
When you call an undeclared function, the C89 standard mandates an implicit declaration. This declaration would be:
int pack_cui();
The ()
is not the same as (void)
, it indicates that the function takes an unspecified number of arguments, whereas (void)
means zero. This is left over from pre-ISO/ANSI C, back in the K&R days.
You don't want that, because that is the wrong declaration. Create your own declaration at the top, above main()
:
unsigned pack_cui(char *);
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