Thread: returning allocated space from function

Hello, this could be a stupid question, but I don't know the answer...a short example explains more than many words:

char* function(){
char *str;

str=(char *)MallocSafe(SIZE);
return str;
}

void main(){
char *buffer;

buffer=function();
}

In this case, will the value of the pointer "buffer" be the same of "str"? In other words, will the memory allocated and pointed to by str be accessible again outside?

I'd like to get a technical answer to understand the behavior of the compiler and linker in this case, thanks for help.

BrownB

Yes, both buffer and str would be pointing to the same memory location.

i'm o.k with you thantos

I'd like to get a technical answer to understand the behavior of the compiler and linker in this case, thanks for help.

@==adresse ...
str is @ of (*str) so your function return the adress not the caracter
buffer receive (@*str) so buffer will have the adress of *str then buffer and str will point the same bloc

Code:

sorry about my english because i use a translator 

>str=(char *)MallocSafe(SIZE);
If you have to cast it, is it really that safe?

>void main(){
main has never been defined to return void, it returns int and nothing else. In other words, your definition should be (assuming you don't want main to take arguments):

Code:

int main(void)

But because this is a definition and not a declaration, this is also acceptable:

Code:

int main()

>In this case, will the value of the pointer "buffer" be the same of "str"?
In this case, no. str and buffer are in mutually exclusive scopes, so by the time execution gets back to buffer, str has been destroyed. However, the memory that str pointed to still exists and is referenced by buffer, so a resource leak isn't caused by losing access to the memory.