function calls during a standard c++ program execution - c++

I have the following code:
int func(int a)
{
int b=2;
int c,d,e,f,g;
//some code which involves a,b,c,d,e,f,g
}
int main()
{
int s=3;
func(s);
}
Now what happens is that when main begins execution:
1.It pushes s onto stack
2.It calls func()
3.func() pushes b,c,d,e,f,g onto stack
4.Now when the code involving a,b,c,d,e,f.g is executed, in order to know the value of a all local variables of func() will have to be popped. Then a's value is retrieved. Now if again b.c.d.e.f.g are to be used, how will their values be retrieved (because they have already been popped)?

The local variables, as well as the argument, aren't actually pushed on the stack. Instead the compiler adds code to change the stack pointer by enough to fit all variables, and then when referencing a local variable the compiler have code to get the value from an offset from the stack pointer.
I recommend that you look at the assembler output of your example program to understand how it works.

The equivalent code for void func(int a)
void func(int a)
{
00413880 push ebp
00413881 mov ebp,esp
00413883 sub esp,108h
00413889 push ebx
0041388A push esi
0041388B push edi
0041388C lea edi,[ebp-108h]
00413892 mov ecx,42h
00413897 mov eax,0CCCCCCCCh
0041389C rep stos dword ptr es:[edi]
int b=2;
0041389E mov dword ptr [b],2
int c,d,e,f,g;
//some code which involves a,b,c,d,e,f,g
}
Now lets see the equivalent assembly code for the below code::
void func(int a)
{
int b=2;
int c,d,e,f,g;
c = 10 ;
d = 15 ;
e = 20 ;
a = a + 2 ;
}
Assembly Code::
void func(int a)
{
00413880 push ebp
00413881 mov ebp,esp
00413883 sub esp,108h
00413889 push ebx
0041388A push esi
0041388B push edi
0041388C lea edi,[ebp-108h]
00413892 mov ecx,42h
00413897 mov eax,0CCCCCCCCh
0041389C rep stos dword ptr es:[edi]
int b=2;
0041389E mov dword ptr [b],2
int c,d,e,f,g;
c = 10 ;
004138A5 mov dword ptr [c],0Ah
d = 15 ;
004138AC mov dword ptr [d],0Fh
e = 20 ;
004138B3 mov dword ptr [e],14h
a = a + 2 ;
004138BA mov eax,dword ptr [a]
004138BD add eax,2
004138C0 mov dword ptr [a],eax
}
So, although they are pushed in Stack (ESP or SP) but the pointer to each of the local variables are stored too, so that, they can be accessed easily when needed.
Like, when the code needed to use variable a, the dword ptr [a] is simply moved to register EAX.
NOTE: Technically not pushed but adjusted to fit in all variables. ( Courtesy: Joachim Pileborg)

in order to know the value of a all local variables of func() will have to be popped
This line is although grammatically ambiguous. I ll assume you mean to say the varibles are popped when being used by function..
But the actual case is local variable are popped out only when function returns to the caller.
Provided they are automatic.
On the other hand when function wants to access(read/write) the variable. It uses the offset(distance) from the base to access them, so there is no question of variables being popped out while being accessed for evaluation.

Related

Do modern compilers optimize local variables that just refer to subset of objects?

Do most modern compilers end up optimizing the following code so that extra instructions aren't used for the object inner?
func Test(TypeObject *object):
InnerTypedObject *inner = object->inner
print(inner->a)
print(inner->b)
print(inner->c)
I figured that compilers would be able to figure out that inner->a and object->inner.a refer to the same thing, so it would avoid allocating inner altogether. I figured the local variable is probably saved on a register, so I'm not really concerned about performance. Mainly wanted to know if we'd get the same generated machine code.
Thanks to Jerry Coffin for the comment - my original answer was actually quite wrong...
For this code:
struct TypeObject {
int a;
int b;
int c;
};
void print(int x);
void test(TypeObject *object) {
print(object->a);
print(object->b);
print(object->c);
}
https://godbolt.org/g/SrNWkp produces something like this:
test(TypeObject*):
push rbx // save the rbx register
mov rbx, rdi // copy the parameter (which is "object") to rbx
mov edi, DWORD PTR [rbx] // copy inner->a to edi
call print(int)
mov edi, DWORD PTR [rbx+4] // copy inner->b to edi
call print(int)
mov edi, DWORD PTR [rbx+8] // copy inner->c to edi
jmp print(int)
pop rbx // restore rbx
And for this code:
struct InnerTypedObject {
int a;
int b;
int c;
};
struct TypeObject {
InnerTypedObject * inner;
};
void print(int x);
void test(TypeObject *object) {
InnerTypedObject *inner = object->inner;
print(inner->a);
print(inner->b);
print(inner->c);
}
https://godbolt.org/g/NC2pa3 produces something like this:
test(TypeObject*):
push rbx // save the rbx register
mov rbx, QWORD PTR [rdi] // copy "*object" (which is "inner") to rbx
mov edi, DWORD PTR [rbx] // copy inner->a to edi
call print(int)
mov edi, DWORD PTR [rbx+4] // copy inner->b to edi
call print(int)
mov edi, DWORD PTR [rbx+8] // copy inner->c to edi
jmp print(int)
pop rbx // restore rbx
So the code is still dereferencing object - it stores the pointer once and then uses it three times just like the original code did. The reason for not being able to optimize it better is that what is stored in a pointer is extremely hard to track so the optimizer has to assume it doesn't know what is in there for sure.
Even though both bits of assembly have the same number of instructions, there is an extra memory dereference in the one with "inner" so it could be expensive if the data isn't already in the cache.

Accessing value using get method or member directly?

class SomeClass
{
int classMember;
public:
int GetMember();
bool IsPositive();
};
int SomeClass::GetMember()
{
return classMember;
}
bool SomeClass::IsPositive()
{
int val = GetMember(); //Case#1. Accessing value using get method
int val = classMember; //Case#2. Accessing value directly
return (val > 0);
}
Question: Does using Case#1 have any advantages over Case#2 or vice versa? Are there any overheads (even the tiniest) in using get method as compared to the member directly?
The answer is, it really depends on how your compiler chooses to do something. The best way to see if there is a difference is to look at the disassembly.
int val = classMember;
010C4869 mov eax,dword ptr [this]
010C486C mov ecx,dword ptr [eax]
010C486E mov dword ptr [val],ecx
return val > 0;
010C4871 cmp dword ptr [val],0
010C4875 jle SomeClass::IsPositiveClassMember+20h (010C4880h)
010C4877 mov dword ptr [ebp-4],1
010C487E jmp SomeClass::IsPositiveClassMember+27h (010C4887h)
010C4880 mov dword ptr [ebp-4],0
010C4887 mov al,byte ptr [ebp-4]
vs.
int val = GetMember();
010C4829 mov ecx,dword ptr [this]
010C482C call SomeClass::GetMember (010C1168h)
010C4831 mov dword ptr [val],eax
return val > 0;
010C4834 cmp dword ptr [val],0
010C4838 jle SomeClass::IsPositiveGetMember+23h (010C4843h)
010C483A mov dword ptr [ebp-4],1
010C4841 jmp SomeClass::IsPositiveGetMember+2Ah (010C484Ah)
010C4843 mov dword ptr [ebp-4],0
010C484A mov al,byte ptr [ebp-4]
The second example calls SomeClass::GetMember, which has its own disassembly. So in the second case, instead of just loading the value from member, it makes a function call.
return classMember;
010C4817 mov eax,dword ptr [this]
010C481A mov eax,dword ptr [eax]
You'll note that the instructions to load val with the value of classMember are identical, so the overhead comes from the Call SomeClass::GetMember.
This is in debug mode however, with no optimization. If we optimize, and build in release, we see the following disassembly:
int val = classMember;
return val > 0;
013D4830 xor eax,eax
013D4832 cmp dword ptr [ecx],eax
013D4834 setg al
vs
int val = GetMember();
return val > 0;
013D4820 xor eax,eax
013D4822 cmp dword ptr [ecx],eax
013D4824 setg al
The compiler optimizes away the call, and there is no difference.
Its purely depend on the need of programmer.
When to create function?
To increase the modularity of the program
When some common tasks are performed repeatedly.
In your case say suppose two number to be added.
For that you may use
int a= data member 1 + data member 2
Say suppose you have to use that at many places.
Then you have consider the point number 2.
You can just create the function like addnumber() this for ease of use and readability.
Regarding performance,
Both are same because the member function are inline by default when defined inside class definition. So separate stack allocation for calling that function is not required
Using GetMember() might be a bit slower, unless its made inline explicitly or implicitly by your compiler.
However, using an accessor can greatly help your debugging, temporarily changing this:
int SomeClass::GetMember()
{
return classMember;
}
into this:
int SomeClass::GetMember()
{
std::cout << "GetMember() called when classMember=" << classMember << std::endl;
return classMember;
}
But this might be a bit oldschool.

classes from c++ to assembly language

I can't understand how classes are implemented in assembly language. I inspected the source code of a c++ application( compiler of visual studio ) that uses a class but it looks like a normal code with no special functions or something other. Where is the constructor of this class and how does it work? I guess the parameters of the constructor are passed by stack, but what does function unknown_libname_1 do?
.text:00261050
.text:00261050 ; int __cdecl main(int argc, const char **argv, const char **envp)
.text:00261050 _main proc near ; CODE XREF: ___tmainCRTStartup+10Ap
.text:00261050
.text:00261050 var_10 = byte ptr -10h
.text:00261050 var_8 = byte ptr -8
.text:00261050 argc = dword ptr 8
.text:00261050 argv = dword ptr 0Ch
.text:00261050 envp = dword ptr 10h
.text:00261050
.text:00261050 push ebp
.text:00261051 mov ebp, esp
.text:00261053 sub esp, 10h
.text:00261056 push 4
.text:00261058 push 3
.text:0026105A lea ecx, [ebp+var_8]
.text:0026105D call unknown_libname_1 ; Microsoft VisualC 2-10/net runtime
.text:00261062 push 6
.text:00261064 push 5
.text:00261066 lea ecx, [ebp+var_10]
.text:00261069 call unknown_libname_1 ; Microsoft VisualC 2-10/net runtime
.text:0026106E mov eax, ds:?endl#std##YAAAV?$basic_ostream#DU?$char_traits#D#std###1#AAV21##Z ; std::endl(std::basic_ostream<char,std::char_traits<char>> &)
.text:00261073 push eax
.text:00261074 lea ecx, [ebp+var_8]
.text:00261077 call sub_261000
.text:0026107C push eax
.text:0026107D push offset aRectArea ; "rect area: "
.text:00261082 mov ecx, ds:?cout#std##3V?$basic_ostream#DU?$char_traits#D#std###1#A ; std::basic_ostream<char,std::char_traits<char>> std::cout
.text:00261088 push ecx
.text:00261089 call sub_2612D0
.text:0026108E add esp, 8
.text:00261091 mov ecx, eax
.text:00261093 call ds:??6?$basic_ostream#DU?$char_traits#D#std###std##QAEAAV01#H#Z ; std::basic_ostream<char,std::char_traits<char>>::operator<<(int)
.text:00261099 mov ecx, eax
.text:0026109B call ds:??6?$basic_ostream#DU?$char_traits#D#std###std##QAEAAV01#P6AAAV01#AAV01##Z#Z ; std::basic_ostream<char,std::char_traits<char>>::operator<<(std::basic_ostream<char,std::char_traits<char>> & (*)(std::basic_ostream<char,std::char_traits<char>> &))
.text:002610A1 mov edx, ds:?endl#std##YAAAV?$basic_ostream#DU?$char_traits#D#std###1#AAV21##Z ; std::endl(std::basic_ostream<char,std::char_traits<char>> &)
.text:002610A7 push edx
.text:002610A8 lea ecx, [ebp+var_10]
.text:002610AB call sub_261000
.text:002610B0 push eax
.text:002610B1 push offset aRectbArea ; "rectb area: "
.text:002610B6 mov eax, ds:?cout#std##3V?$basic_ostream#DU?$char_traits#D#std###1#A ; std::basic_ostream<char,std::char_traits<char>> std::cout
.text:002610BB push eax
.text:002610BC call sub_2612D0
.text:002610C1 add esp, 8
.text:002610C4 mov ecx, eax
.text:002610C6 call ds:??6?$basic_ostream#DU?$char_traits#D#std###std##QAEAAV01#H#Z ; std::basic_ostream<char,std::char_traits<char>>::operator<<(int)
.text:002610CC mov ecx, eax
.text:002610CE call ds:??6?$basic_ostream#DU?$char_traits#D#std###std##QAEAAV01#P6AAAV01#AAV01##Z#Z ; std::basic_ostream<char,std::char_traits<char>>::operator<<(std::basic_ostream<char,std::char_traits<char>> & (*)(std::basic_ostream<char,std::char_traits<char>> &))
.text:002610D4 xor eax, eax
.text:002610D6 mov esp, ebp
.text:002610D8 pop ebp
.text:002610D9 retn
.text:0026
function sub_261000:
sub_261000 proc near
var_4= dword ptr -4
push ebp
mov ebp, esp
push ecx
mov [ebp+var_4], ecx
mov eax, [ebp+var_4]
mov ecx, [ebp+var_4]
mov eax, [eax]
imul eax, [ecx+4]
mov esp, ebp
pop ebp
retn
sub_261000 endp
Assembly doesn't have any concept of classes. When you create an object of class type on the stack, it just makes room for all of its member objects. A member function is just like a normal function, but it is also passed a pointer to the beginning of these member objects, which is the this pointer. The compiled functions just access members relative to this pointer. The constructor is just another function that initialises these member objects.
You could think of this:
class A {
private:
int x;
short y;
public:
A(int arg) : x(arg), y(6) { }
void print() {
std::cout << x << ',' << y << std::endl;
}
};
int main() {
A a(5);
a.print()
}
As being transformed into something like this made-up invalid C++:
void A_construct(A* this, int arg) {
this->x = arg;
this->y = 6;
}
void A_print(A* this) {
std::cout << this->x << ',' << this->y << std::endl;
}
int main() {
int x;
short y;
A_construct(this_cast<A*>(&x), 5);
A_print(this_cast<A*>(&x));
}
The only reason the made-up this_cast is there is to allow the A* so I can continue to use this->x to mean "Access the x object of that A". Just a convenience for illustration.
Other than these run-time details, the only other effect that classes have is that they place some compile-time restrictions on your code. For example, you can't write code that access a private member of a class from outside of it. That's not enforced in the assembly in any way.
I have not really an idea what you are searching for :-)
But at first you have to keep in mind:
1) the compiler will try to optimize all your code, independent of your language. For that you will typically see no inlined functions at all.
2) All calculations from constant values will be done in compile time as part of optimization. So you may only the the processing of the values if they are not compile time constant!
3) Initialization of global objects will be done by a simple copy loop at start of your program which is running before main. There is nothing to see from the code itself.
4) You can only see a data structure of class. But here you have to think a about padding and other things.
Sometimes it is a good idea to look in the produced assembler code to get a feeling which type of programming results in which kind of executable. But today most compilers do a really perfect job which makes it nearly impossible to see how the code was presented in the high level language. But yes, it is good for learning :-)

How to use variables in __asm?

I'm compiling this C++ code with the VC compiler. I'm trying to call a function that takes two WORD (aka unsigned short) parameters using the __asm statement, like this:
__declspec(naked) void __stdcall CallFunction(WORD a, WORD b)
{
__asm {
PUSH EBP
MOV EBP, ESP
PUSH a
PUSH b
CALL functionAddress
LEAVE
RETN
}
}
The function at functionAddress simply outputs the result of doing a + b. Then calling CallFuncion(5, 5); prints "64351" or something like that. The problem is when using the a and b variables inside the __asm statement because this works:
PUSH EBP
MOV EBP, ESP
PUSH 5
PUSH 5
CALL functionAddress
LEAVE
This is the function at functionAddress:
void __stdcall Add(WORD a, WORD b)
{
WORD c;
c = a + b;
printf("The result is %d\n", c);
}
How can I do this the right way? So the __asm statement interpretate the a and b values?
Since you're using __declspec(naked) and setting up your own stack frame, I don't believe the compiler will let you refer to a and b by name. Using __declspec(naked) basically means you're responsible for dealing with the stack frame, parameters, etc., on your own.
You probably want code more on this general order:
__asm {
PUSH EBP
MOV EBP, ESP
mov eax, [ebp+8]
mov ebx, [ebp+12]
push eax
push ebx
CALL functionAddress
LEAVE
RETN
}
I'ts been a while since I've handled things like this by hand, so you might want to re-check those offsets, but if I recall correctly, the return address should be at [ebp+4]. Parameters are (usually) pushed from right to left, so the the left-most parameter should be next at [ebp+8], and the next parameter at [ebp+12] (keeping in mind that the stack grows downward).
Edit: [I should have looked more carefully at the function heading.]
You've marked CallFunction as using the __stdcall calling convention. That means it's required to clean up the parameters that were passed to it. So, since it receives 8 bytes of parameters, it needs to remove 8 bytes from the stack as it returns:
PUSH EBP
MOV EBP, ESP
mov eax, [ebp+8]
mov ebx, [ebp+12]
push eax
push ebx
CALL Add_f
LEAVE
RET 8

How does returning values from a function work?

I recently had a serious bug, where I forgot to return a value in a function. The problem was that even though nothing was returned it worked fine under Linux/Windows and only crashed under Mac. I discovered the bug when I turned on all compiler warnings.
So here is a simple example:
#include <iostream>
class A{
public:
A(int p1, int p2, int p3): v1(p1), v2(p2), v3(p3)
{
}
int v1;
int v2;
int v3;
};
A* getA(){
A* p = new A(1,2,3);
// return p;
}
int main(){
A* a = getA();
std::cerr << "A: v1=" << a->v1 << " v2=" << a->v2 << " v3=" << a->v3 << std::endl;
return 0;
}
My question is how can this work under Linux/Windows without crashing? How is the returning of values done on lower level?
On Intel architecture, simple values (integers and pointers) are usually returned in eax register. This register (among others) is also used as temporary storage when moving values in memory and as operand during calculations. So whatever value left in that register is treated as the return value, and in your case it turned out to be exactly what you wanted to be returned.
Probably by luck, 'a' left in a register that happens to be used for returning single pointer results, something like that.
The calling/ conventions and function result returns are architecture-dependent, so it's not surprising that your code works on Windows/Linux but not on a Mac.
There are two major ways for a compiler to return a value:
Put a value in a register before returning, and
Have the caller pass a block of stack memory for the return value, and write the value into that block [more info]
The #1 is usually used with anything that fits into a register; #2 is for everything else (large structs, arrays, et cetera).
In your case, the compiler uses #1 both for the return of new and for the return of your function. On Linux and Windows, the compiler did not perform any value-distorting operations on the register with the returned value between writing it into the pointer variable and returning from your function; on Mac, it did. Hence the difference in the results that you see: in the first case, the left-over value in the return register happened to co-inside with the value that you wanted to return anyway.
First off, you need to slightly modify your example to get it to compile. The function must have at least an execution path that returns a value.
A* getA(){
if(false)
return NULL;
A* p = new A(1,2,3);
// return p;
}
Second, it's obviously undefined behavior, which means anything can happen, but I guess this answer won't satisfy you.
Third, in Windows it works in Debug mode, but if you compile under Release, it doesn't.
The following is compiled under Debug:
A* p = new A(1,2,3);
00021535 push 0Ch
00021537 call operator new (211FEh)
0002153C add esp,4
0002153F mov dword ptr [ebp-0E0h],eax
00021545 mov dword ptr [ebp-4],0
0002154C cmp dword ptr [ebp-0E0h],0
00021553 je getA+7Eh (2156Eh)
00021555 push 3
00021557 push 2
00021559 push 1
0002155B mov ecx,dword ptr [ebp-0E0h]
00021561 call A::A (21271h)
00021566 mov dword ptr [ebp-0F4h],eax
0002156C jmp getA+88h (21578h)
0002156E mov dword ptr [ebp-0F4h],0
00021578 mov eax,dword ptr [ebp-0F4h]
0002157E mov dword ptr [ebp-0ECh],eax
00021584 mov dword ptr [ebp-4],0FFFFFFFFh
0002158B mov ecx,dword ptr [ebp-0ECh]
00021591 mov dword ptr [ebp-14h],ecx
The second instruction, the call to operator new, moves into eax the pointer to the newly created instance.
A* a = getA();
0010484E call getA (1012ADh)
00104853 mov dword ptr [a],eax
The calling context expects eax to contain the returned value, but it does not, it contains the last pointer allocated by new, which is incidentally, p.
So that's why it works.
As Kerrek SB mentioned, your code has ventured into the realm of undefined behavior.
Basically, your code is going to compile down to assembly. In assembly, there's no concept of a function requiring a return type, there's just an expectation. I'm the most comfortable with MIPS, so I shall use MIPS to illustrate.
Assume you have the following code:
int add(x, y)
{
return x + y;
}
This is going to be translated to something like:
add:
add $v0, $a0, $a1 #add $a0 and $a1 and store it in $v0
jr $ra #jump back to where ever this code was jumped to from
To add 5 and 4, the code would be called something like:
addi $a0, $0, 5 # 5 is the first param
addi $a1, $0, 4 # 4 is the second param
jal add
# $v0 now contains 9
Note that unlike C, there's no explicit requirement that $v0 contain the return value, just an expectation. So, what happens if you don't actually push anything into $v0? Well, $v0 always has some value, so the value will be whatever it last was.
Note: This post makes some simplifications. Also, you're computer is likely not running MIPS... But hopefully the example holds, and if you learned assembly at a university, MIPS might be what you know anyway.
The way of returning of value from the function depends on architecture and the type of value. It could be done thru registers or thru stack.
Typically in the x86 architecture the value is returned in EAX register if it is an integral type: char, int or pointer.
When you don't specify the return value, that value is undefined. This is only your luck that your code sometimes worked correctly.
When popping values from the stack in IBM PC architecture there is no physical destruction of the old values ​​of data stored there. They just become unavailable through the operation of the stack, but still remain in the same memory cell.
Of course, the previous values ​​of these data will be destroyed during the subsequent pushing of new data on the stack.
So probably you are just lucky enough, and nothing is added to stack during your function's call and return surrounding code.
Regarding the following statement from n3242 draft C++ Standard, paragraph 6.6.3.2, your example yields undefined behavior:
Flowing off the end of a function is equivalent to a return with no
value; this results in undefined behavior in a value-returning
function.
The best way to see what actually happens is to check the assembly code generated by the given compiler on a given architecture. For the following code:
#pragma warning(default:4716)
int foo(int a, int b)
{
int c = a + b;
}
int main()
{
int n = foo(1, 2);
}
...VS2010 compiler (in Debug mode, on Intel 32-bit machine) generates the following assembly:
#pragma warning(default:4716)
int foo(int a, int b)
{
011C1490 push ebp
011C1491 mov ebp,esp
011C1493 sub esp,0CCh
011C1499 push ebx
011C149A push esi
011C149B push edi
011C149C lea edi,[ebp-0CCh]
011C14A2 mov ecx,33h
011C14A7 mov eax,0CCCCCCCCh
011C14AC rep stos dword ptr es:[edi]
int c = a + b;
011C14AE mov eax,dword ptr [a]
011C14B1 add eax,dword ptr [b]
011C14B4 mov dword ptr [c],eax
}
...
int main()
{
011C14D0 push ebp
011C14D1 mov ebp,esp
011C14D3 sub esp,0CCh
011C14D9 push ebx
011C14DA push esi
011C14DB push edi
011C14DC lea edi,[ebp-0CCh]
011C14E2 mov ecx,33h
011C14E7 mov eax,0CCCCCCCCh
011C14EC rep stos dword ptr es:[edi]
int n = foo(1, 2);
011C14EE push 2
011C14F0 push 1
011C14F2 call foo (11C1122h)
011C14F7 add esp,8
011C14FA mov dword ptr [n],eax
}
The result of addition operation in foo() is stored in eax register (accumulator) and its content is used as a return value of the function, moved to variable n.
eax is used to store a return value (pointer) in the following example as well:
#pragma warning(default:4716)
int* foo(int a)
{
int* p = new int(a);
}
int main()
{
int* pn = foo(1);
if(pn)
{
int n = *pn;
delete pn;
}
}
Assembly code:
#pragma warning(default:4716)
int* foo(int a)
{
000C1520 push ebp
000C1521 mov ebp,esp
000C1523 sub esp,0DCh
000C1529 push ebx
000C152A push esi
000C152B push edi
000C152C lea edi,[ebp-0DCh]
000C1532 mov ecx,37h
000C1537 mov eax,0CCCCCCCCh
000C153C rep stos dword ptr es:[edi]
int* p = new int(a);
000C153E push 4
000C1540 call operator new (0C1253h)
000C1545 add esp,4
000C1548 mov dword ptr [ebp-0D4h],eax
000C154E cmp dword ptr [ebp-0D4h],0
000C1555 je foo+50h (0C1570h)
000C1557 mov eax,dword ptr [ebp-0D4h]
000C155D mov ecx,dword ptr [a]
000C1560 mov dword ptr [eax],ecx
000C1562 mov edx,dword ptr [ebp-0D4h]
000C1568 mov dword ptr [ebp-0DCh],edx
000C156E jmp foo+5Ah (0C157Ah)
std::operator<<<std::char_traits<char> >:
000C1570 mov dword ptr [ebp-0DCh],0
000C157A mov eax,dword ptr [ebp-0DCh]
000C1580 mov dword ptr [p],eax
}
...
int main()
{
000C1610 push ebp
000C1611 mov ebp,esp
000C1613 sub esp,0E4h
000C1619 push ebx
000C161A push esi
000C161B push edi
000C161C lea edi,[ebp-0E4h]
000C1622 mov ecx,39h
000C1627 mov eax,0CCCCCCCCh
000C162C rep stos dword ptr es:[edi]
int* pn = foo(1);
000C162E push 1
000C1630 call foo (0C124Eh)
000C1635 add esp,4
000C1638 mov dword ptr [pn],eax
if(pn)
000C163B cmp dword ptr [pn],0
000C163F je main+51h (0C1661h)
{
int n = *pn;
000C1641 mov eax,dword ptr [pn]
000C1644 mov ecx,dword ptr [eax]
000C1646 mov dword ptr [n],ecx
delete pn;
000C1649 mov eax,dword ptr [pn]
000C164C mov dword ptr [ebp-0E0h],eax
000C1652 mov ecx,dword ptr [ebp-0E0h]
000C1658 push ecx
000C1659 call operator delete (0C1249h)
000C165E add esp,4
}
}
VS2010 compiler issues warning 4716 in both examples. By default this warning is promoted to an error.