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这个问题在这里已有答案：
- Initializing a pointer in a separate function in C 2 answers
- 在C 2答案中的单独函数中初始化指针
i have a file function.c and main.c. In function.c there is this function
我有一个文件function.c和main.c.在function.c中有这个功能
int GetRow(int descriptor,char* src)
{
char app[1];
char* carattere= (char*)malloc(sizeof(char));
int count=0;
int capp=0;
while((capp=read(descriptor,app,sizeof(char)))>0
&&app[0]!=';')
{
if(app[0]!='\n')
{
carattere[count]=app[0];
carattere=(char*)realloc(carattere,sizeof(char)*(++count +1));
}
}
src=carattere;
if(capp<0)
{
return -1;
};
#define DEBUG
#ifdef DEBUG
printf("The line Detected was %s %s\n",carattere,src);
#endif
}
And it work becouse when i use printf to see if the src point to the new address the return the same thing. The problem born in main.c where i call GetRow
它工作因为当我使用printf来查看src是否指向新地址返回相同的东西。问题出现在main.c中，我称之为GetRow
char* pointer;
int file=open("prova.conf",O_RDWR);
GetRow(file,pointer);
printf("%s",pointer);
Becouse when i use printf itprint null. Using gdb then the call to GetRow i understand that the pointer point 0x0, so please can anyone tell and explain my issue?? Thanks and excuse me for my english.
因为我使用printf itprint null。使用gdb然后调用GetRow我明白指针指向0x0，所以请任何人可以告诉和解释我的问题??谢谢，请原谅我的英语。
1 个解决方案
#1
2
You still passing the char * src by value. If you want to change the value of the pointer you need to pass a reference to it. use char **src and set *src = carattere;
您仍然按值传递char * src。如果要更改指针的值，则需要传递对它的引用。使用char ** src并设置* src = carattere;
Just because you're passing a pointer doesn't mean you're necessarily passing by reference. If you malloc memory for src in main and then pass the reference to that memory (as you have the char * src) you can change the value at that reference by *src = *carattere but that's probably not what you want either.
仅仅因为你传递一个指针并不意味着你必须通过引用传递。如果你在main中为src malloc内存然后将引用传递给那个内存（因为你有char * src），你可以用* src = * carattere更改该引用的值，但这可能不是你想要的。
#1
2
You still passing the char * src by value. If you want to change the value of the pointer you need to pass a reference to it. use char **src and set *src = carattere;
您仍然按值传递char * src。如果要更改指针的值，则需要传递对它的引用。使用char ** src并设置* src = carattere;
Just because you're passing a pointer doesn't mean you're necessarily passing by reference. If you malloc memory for src in main and then pass the reference to that memory (as you have the char * src) you can change the value at that reference by *src = *carattere but that's probably not what you want either.
仅仅因为你传递一个指针并不意味着你必须通过引用传递。如果你在main中为src malloc内存然后将引用传递给那个内存（因为你有char * src），你可以用* src = * carattere更改该引用的值，但这可能不是你想要的。