stack increase direction and big-endian or little-endia

[gary]

Hi,

1. About all C/C++ compilers, Does stack increase from high address to low address and heap grow increase from low to high? What on earth decides their increase direction, CPU architecture, OS or compiler?

2. In GNU gcc,
{
int a[] = {0, 1, 2, 3, 4};
bool b;
float c;
for (int i = 0; i < 5; i++) {
cout << hex << a+i << "," ;
}
cout << endl;
cout << hex << &a << "," << &b << "," << &c << endl;
}
one print result is: 0x22ff40, 0x22ff44, 0x22ff48, 0x22ff4c, 0x22ff50
0x22ff40, 0x22ff3f, 0x22ff38
So considering question one, why the address of the array(a) increased to high address?


3. What decides the big-endian and little-endian, CPU architecture, OS or compiler?

4. {
int x = 0x12345678;
char *cp = (char*)&x;
for(int i = 0; i < 4; i++) {
cout << hex << (int)*(cp+i) << " ";
}
}
what is run result?
I think it depends on two aspects: stack increase direction and big-endian or little-endia.
Considering in the case of big-endian, suppose the address of x is 0x22ff44, namely &x=0x22ff44, and sizeof(x)=4.
If stack increases to low address, the variable x occupied 0x22ff44, 0x22ff43, 0x22ff42 and 0x22ff41, so the result is "78 56 34 12".
If stack increases to high address, the variable x occupied 0x22ff44, 0x22ff45, 0x22ff46 and 0x22ff47, so the result is "12 34 56 78".

Please give me some advice, thank you.

 

[Andrej Hristoliubov]

Hi,

1. About all C/C++ compilers, Does stack increase from high address to low address and heap grow increase from low to high? What on earth decides their increase direction, CPU architecture, OS or compiler?

2. In GNU gcc,
{
int a[] = {0, 1, 2, 3, 4};
bool b;
float c;
for (int i = 0; i < 5; i++) {
cout << hex << a+i << "," ;
}
cout << endl;
cout << hex << &a << "," << &b << "," << &c << endl;
}
one print result is: 0x22ff40, 0x22ff44, 0x22ff48, 0x22ff4c, 0x22ff50
0x22ff40, 0x22ff3f, 0x22ff38
So considering question one, why the address of the array(a) increased to high address?


3. What decides the big-endian and little-endian, CPU architecture, OS or compiler?

4. {
int x = 0x12345678;
char *cp = (char*)&x;
for(int i = 0; i < 4; i++) {
cout << hex << (int)*(cp+i) << " ";
}
}
what is run result?
I think it depends on two aspects: stack increase direction and big-endian or little-endia.
Considering in the case of big-endian, suppose the address of x is 0x22ff44, namely &x=0x22ff44, and sizeof(x)=4.
If stack increases to low address, the variable x occupied 0x22ff44, 0x22ff43, 0x22ff42 and 0x22ff41, so the result is "78 56 34 12".
If stack increases to high address, the variable x occupied 0x22ff44, 0x22ff45, 0x22ff46 and 0x22ff47, so the result is "12 34 56 78".

Please give me some advice, thank you.

all three

 

[Greg]

Hi,

1. About all C/C++ compilers, Does stack increase from high address to low address and heap grow increase from low to high? What on earth decides their increase direction, CPU architecture, OS or compiler?

The direction that the stack and heap grows is OS-dependent and has
nothing to do with C or C++ compilers. Generally, one can expect that
the heap and the stack will grow in opposite directions and from
opposite ends of the logical address space (in order that they not
collide unnecessarily).

2. In GNU gcc,
{
int a[] = {0, 1, 2, 3, 4};
bool b;
float c;
for (int i = 0; i < 5; i++) {
cout << hex << a+i << "," ;
}
cout << endl;
cout << hex << &a << "," << &b << "," << &c << endl;
}
one print result is: 0x22ff40, 0x22ff44, 0x22ff48, 0x22ff4c, 0x22ff50
0x22ff40, 0x22ff3f, 0x22ff38
So considering question one, why the address of the array(a) increased to high address?

No matter which direction the heap or stack may grow, objects in
memory, including arrays will have the same layout. Each element in an
array will stored at a higher memory location than the element before
it.

3. What decides the big-endian and little-endian, CPU architecture, OS or compiler?

The CPU may only support one kind of endianness, or the OS may be
written for a particular endian-ness. Of course endianness is just a
convention for storing multibyte data. As such, a program is free to
use either convention (or make up its own) with its own data. Generally
in the interests of inter-operabillity and consistency, it is best to
use the same endianess that the OS prefers.

4. {
int x = 0x12345678;
char *cp = (char*)&x;
for(int i = 0; i < 4; i++) {
cout << hex << (int)*(cp+i) << " ";
}
}
what is run result?

On a big endian machine: 12 34 56 78
little-endian: 78 56 34 21

I think it depends on two aspects: stack increase direction and big-endian or little-endia.

The direction the stack increases has no effect on how x would be
stored in memory.

Considering in the case of big-endian, suppose the address of x is 0x22ff44, namely &x=0x22ff44, and sizeof(x)=4.
If stack increases to low address, the variable x occupied 0x22ff44, 0x22ff43, 0x22ff42 and 0x22ff41, so the result is "78 56 34 12".
If stack increases to high address, the variable x occupied 0x22ff44, 0x22ff45, 0x22ff46 and 0x22ff47, so the result is "12 34 56 78".

No, only the endianness used to store x would affect the order in which
its bytes are arranged in memory.

Greg

 

[EventHelix.com]

1. About all C/C++ compilers, Does stack increase from high address to low address and heap grow increase from low to high? What on earth >decides their increase direction, CPU architecture, OS or compiler?

In most cases, it is the CPU architecture. In some cases where the
processor allows the stack to grow in either direction, the compiler
might provide a switch to choose between the two.

2. In GNU gcc,
{
int a[] = {0, 1, 2, 3, 4};
...
So considering question one, why the address of the array(a) increased to high address?

The space for the entire array is allocated in one stroke. Individual
array elements are not pushed on the stack. Note that C arrays operate
the same way regardless of the variables location on stack, heap or
global memory.

3. What decides the big-endian and little-endian, CPU architecture, OS or compiler?

In most cases, it is the CPU architecture. In some cases where the
processor supports little as well as big endian operation, the compiler
might provide a switch to choose between the two.

The following article should help:

http://www.eventhelix.com/RealtimeMantra/ByteAlignmentAndOrdering.htm