Question about return by reference?
in [Visual C]
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From: Daniel Mark on 10 Apr 2006 14:49 Hello all: When I compile const char* msg_to_usr2(void) { const int rsp_cnt = 5; // this is wrong char usr_msgs[] = "Hello"; return usr_msgs; } warning C4172: returning address of local variable or temporary while the following code is correct. const char* msg_to_usr2(void) { const int rsp_cnt = 5; // this is correct char* usr_msgs[] = {"hello"}; return usr_msgs[0]; } Because the syste has assigned a memeory space to usr_msgs[0] which is used to store the memory location of "hello" in heap memory, the program returns the string without error. I guess there are some other reasons for it. May anyone give me some comments? THank you -Daniel
From: William DePalo [MVP VC++] on 10 Apr 2006 14:57 "Daniel Mark" <daneilmarkhot(a)hotmail.com> wrote in message news:OGPAP%23MXGHA.4620(a)TK2MSFTNGP04.phx.gbl... > Hello all: > > When I compile > const char* msg_to_usr2(void) > { > const int rsp_cnt = 5; > > // this is wrong > char usr_msgs[] = "Hello"; > return usr_msgs; > } > > warning C4172: returning address of local variable or temporary > > while the following code is correct. > const char* msg_to_usr2(void) > { > const int rsp_cnt = 5; > // this is correct > char* usr_msgs[] = {"hello"}; > return usr_msgs[0]; > } > > Because the syste has assigned a memeory space to usr_msgs[0] which is > used to store > the memory location of "hello" in heap memory, the program returns the > string without error. > > I guess there are some other reasons for it. > > > May anyone give me some comments? In the first snippet, you are trying to return the _address_ of something on the stack. And as the stack will be "popped" of its local variables on the return instruction, that's a problem. In the second snippet, you are returning a single byte to the caller. No problem here because that byte is copied to the return value passed by the caller (it fits easily into a register in x86 environment) Regards, Will
From: Doug Harrison [MVP] on 10 Apr 2006 15:10 On Mon, 10 Apr 2006 13:49:52 -0500, "Daniel Mark" <daneilmarkhot(a)hotmail.com> wrote: >Hello all: > >When I compile >const char* msg_to_usr2(void) >{ > const int rsp_cnt = 5; > >// this is wrong > char usr_msgs[] = "Hello"; > return usr_msgs; >} > >warning C4172: returning address of local variable or temporary Could be fixed by making usr_msgs static. >while the following code is correct. >const char* msg_to_usr2(void) >{ > const int rsp_cnt = 5; >// this is correct > char* usr_msgs[] = {"hello"}; > return usr_msgs[0]; >} > >Because the syste has assigned a memeory space to usr_msgs[0] which is used >to store >the memory location of "hello" in heap memory, the program returns the >string without error. > >I guess there are some other reasons for it. > > >May anyone give me some comments? The second one returns the address of a string literal, and string literals exist throughout a program's lifetime. That's the only reason it works. The fact that usr_msgs is involved is incidental, as you're returning a copy of the contents of usr_msgs[0]; that is, the array usr_msgs needn't exist after the function returns. It would be incorrect if you had written: char* f() { char c; char* usr_msgs[] = { &c }; return usr_msgs[0]; // Bad } This one could be fixed by making "c" static, but again, it doesn't matter that usr_msgs ceases to exist once the function returns. The problem is that the local variable "c" has ceased to exist. -- Doug Harrison Visual C++ MVP
From: William DePalo [MVP VC++] on 10 Apr 2006 15:19 "William DePalo [MVP VC++]" <willd.no.spam(a)mvps.org> wrote in message news:OvkCzCNXGHA.4768(a)TK2MSFTNGP05.phx.gbl... > In the second snippet, you are returning a single byte to the caller. No > problem here because that byte is copied to the return value passed by the > caller (it fits easily into a register in x86 environment) Oops. I missed a level of indirection in the second snippet (I read char* usr_msgs[] = {"hello"}; as char usr_msgs[] = {"hello"}; ). See Doug's reply for the correct answer. Mea culpa. Regards, Will
From: Daniel Mark on 10 Apr 2006 16:19
Hello Doug: I have figured out those problems by following your comments. > char* f() > { > char c; > char* usr_msgs[] = { &c }; > return usr_msgs[0]; // Bad > } > > This one could be fixed by making "c" static, but again, it doesn't matter > that usr_msgs ceases to exist once the function returns. The problem is > that the local variable "c" has ceased to exist. I use MS VC6 to compile the above code without warning(pretty strange) and the output from the screen tells me that this usage is bad. Originally, I expect MS VC6 could give me some compile warning. Many thinks to you! -Daniel Also thanks Will! |