#include<stdio.h>
int main()
{
int a,b;
float e;
char f;
printf("int &a = %u\n",&a);
printf("int &b = %u\n",&b);
printf("float &e = %u\n",&e);
printf("char &f = %u\n",&f);
}
The Output is
int &a = 2293324
int &b = 2293320
float &e = 2293316
char &f = 2293315
But when i use this code and replace the printf for float--
#include<stdio.h>
int main()
{
int a,b;
float e;
char f;
printf("int &a = %u\n",&a);
printf("int &b = %u\n",&b);
printf("char &f = %u\n",&f);
}
Then the Output is
int &a = 2293324
int &b = 2293320
char &f = 2293319
here address is not provided to float, but it is declared on top.
My questions are
Is memory not allocated to variables not used in program?
Why addresses allocated in decreasing order. ex- it's going from 2293324 to 2293320?
1) Is memory not allocated to variables not used in program?
Yes that can happen, the compiler is allowed to optimize it out.
2) Why addresses allocated in decreasing order. ex- it's going from 2293324 to 2293320?
That is usual for most local storage implementations, that they use the CPU supported stack pointer going from stack top to stack bottom. All those local variables will be allocated at the stack most probably.
1) Is memory not allocated to variables not used in program?
It's an allowed optimization; if an unused variable doesn't affect the program's observable behavior, a compiler may just discard it completely. Note that most modern compilers will warn you about unused variables (so you can either remove them from the code or do something with them).
2) Why addresses allocated in decreasing order. ex- it's going from 2293324 to 2293320?
The compiler is not required to allocate storage for separate objects in any particular order, so don't assume that your variables will be allocated in the order they were declared. Also, remember that on x86 and some other systems, the stack grows "downwards" towards decreasing addresses. Remember the top of any stack is simply the location where something was most recently pushed - it has nothing to do with relative address values.
While not specifically required by the standard, local variables are universally located on the program stack.
When you enter a function, one of the first thing done is to decrement the stack pointer to provide space for the local variables.
SUBL #SOMETHING, SP
Where SOMETHING is the amount of space required and SP is the stack pointer register.. In your first example, SOMETHING is probably 13. Then the address of:
f is 0(SP)
e is 1(sp)
b is 5(sp)
a is 9(sp)
I am assuming your compiler did not align the stack pointer. Often they do giving something more like:
f is 3(SP)
e is 4(sp)
b is 8(sp)
a is 12(sp)
And SOMETHING would be rounded up to 16 on a 32-bit system.
You might want to generate an assembly listing using your compiler to see what is going on underneath.
Is memory not allocated to variables not used in program?
Note that for local variable memory is not really allocated. A variable is temporarily bound to a location on the program stack (stack is not required by the standard but is how it is done in most cases). That is why the variable's initial value is undefined. It could have been bound to something else previously.
The compiler does not need to reserve space for variables that are not used. They can be optimized away. Usually, there are compiler settings to instruct not to do this for debugging.
Why addresses allocated in decreasing order. ex- it's going from 2293324 to 2293320?
Program stacks generally grow downward. Starting ye olde days, the program would be at the bottom of the address space, the heap above that and the stack at the opposite end.
The heap would grow towards higher addresses. The stack would grow towards the heap (lower addresses).
While the address spaces can be more complicated than that these days, the downward growth of stacks has stayed.
There is no particular requirement that the compiler map the variables to the stack in descending order but there's a 50/50 chance it will do it that way.
Related
This question already has answers here:
Order of local variable allocation on the stack
(10 answers)
Closed 1 year ago.
I wonder how memory is allocated in c. In the example below, it looks like the compiler allocates some memory for the program and then goes backward. How and why does it work like that?
Code:
int main (void) {
int a;
int b;
int c, d;
printf("a: %p\nb: %p\nc: %p\nd: %p\n", &a, &b, &c, &d);
}
Output:
a: 0x7ffff275351c
b: 0x7ffff2753518
c: 0x7ffff2753514
d: 0x7ffff2753510
This depends on the architecture, but on many systems local variables are stored on the stack. For the compiler it is not strictly necessary to populate the stack top down, so this is depending on the compiler implementation. Usually the compiler saves enough space on the stack and organizes it, as it sees fit.
The compiler may not even put it in memory, but can place the values directly in registers instead, as long as the observable behavior stays the same.
This strongly depends on the specific implementation. The language itself only specifies minimal rules for where and how memory is allocated for different objects (see the C 2011 Online Draft, section 6.2.4., for the complete set of rules).
In most implementations, auto variables (i.e., "local" variables) are created on the stack. While implementations are not required to create variables in the same order in which they are declared, it's not uncommon for them to do so.
On x86/x86-64, the stack grows "downwards" towards decreasing addresses, so each item pushed onto the stack will have a lower address than the previous item.
this is the code :
#include <stdio.h>
#include <stdlib.h>
int main() {
int a = 10;
int b = 20;
//printf("\n&a value %p", &a);
int* x = &b;
x = x + 1;
*x = 5;
printf("\nb value %d", b);
printf("\na value %d", a);
}
I want override a with b adress for test the c overflow but when I comment the line 5(printf fuction) I can't write five in a. While if I print the a adress I can write five in a.
Why?
Sorry for my english and thank you.
The reason this occurred is that all normal compilers store objects with automatic storage duration (objects declared inside a block that are not static or extern) on a stack. Your compiler “pushed” a onto the stack, which means it wrote a to the memory location where the stack pointer was pointing and then decremented the pointer. (Decrementing the pointer adds to the stack, because the stack grows in the direction of decreasing memory addresses. Stacks can be oriented in the other direction, but the behavior you observed strongly suggests your system uses the common direction of growing downward.) Then your compiler pushed b onto the stack. So b ended up at a memory address just below a.
When you took the address of b and added one, that produced the memory address where a is. When you used that address to assign 5, that value was written to where a is.
None of this behavior is defined by the C standard. It is a consequence of the particular compiler you used and the switches you compiled with.
You probably compiled with little or no optimization. With optimization turned on, many compilers would simplify the code by removing unnecessary steps (essentially replacing them with shortcuts), so that 20 and 10 are not actually stored on the stack. A possible result with optimization is that “20” and “10” are printed, and your assignment to *x has no effect. However, the C standard does not say what the behavior must be when you use *x in this way, so the results are determined only by the particular compiler you are using, along with the input switches you give it.
After x = x + 1;, x contains an address that you do not own. And by doing *x = 5; you are trying to write to some location that might not be accessible to you. Thus causing UB. Nothing more can be reasoned about.
I writing a simple program, after declaring a variables I am checking the addresses of variables but memory is not allocated contiguously there are gaps in between.
here is my program. why is leaving gaps i am not understanding
#include < stdio.h >
#include < stdint.h >
int main()
{
char char_one,char_two;
int a = 5,b = 7,*ptr,*ptr_one;
static int *sum_ptr;
printf("address of a %u\n",&a);
printf("address of variable b %u\n",&b);
printf("address of ptr variable %u\n",&ptr);
printf("address of ptr_one variable %u\n",&ptr_one);
printf("address of char_one var %u\n",&char_one);
printf("address of char_two var %u\n",&char_two);
return 0;
}
output:
address of a 2636128020
address of variable b 2636128024
address of ptr variable 2636128000
address of ptr_one variable 2636128008
address of char_one var 2636128030
address of char_two var 2636128031
The C standard does not require memory for variables to be allocated contiguously. In fact, memory might not even be allocated at all if the compiler decides it can optimize by keeping a value in a register instead.
If you declare a struct, the contents of the struct will be ordered the way you declare them, but you still might need to consider how the data is aligned within the struct - for example, ints are aligned on 4-byte boundaries in many architectures. So if you have:
struct foo
{
char a;
int b;
}
a is guaranteed to come before b in memory, but they will still be padded with extra bytes between them to keep the correct alignment (so it takes 8 bytes to store your struct, even though it only "really" needs 5).
Here's a good resource for how structure alignment works:
The lost art of C structure packing
It is not necessary that compiler will allocate memory contiguously (excluding array, In case of structures may be padding there). It is free to allocate memory from where it finds a free location.
This is by design and on purpose, even on a freshly booted machine you won't get contiguous malloc()s.
Read here: http://en.wikipedia.org/wiki/Address_space_layout_randomization
Upon request: address space layout randomization is a techique to make sure memory is allocated in a more or less unpredictable way to ensure attacks relying on a fixed address layout won't succeed. This makes it (more) difficult for an attacker to exploit known deficiencies (or even detect them first).
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Malloc or normal array definition?
We learn that there is dynamic memory in C and dynamic variables:
#include <stdio.h>
int a = 17;
int main(void)
{
int b = 18; //automatic stack memory
int * c;
c = malloc( sizeof( int ) ); //dynamic heap memory
*c = 19;
printf("a = %d at address %x\n", a, &a);
printf("b = %d at address %x\n", b, &b);
printf("c = %d at address %x\n", *c, c);
free(c);
system("PAUSE");
return 0;
}
How do I know which type of memory to use? When do I ned one or the other?
Use dynamic in the following situations:
When you need a lot of memory. Typical stack size is 1 MB, so anything bigger than 50-100KB should better be dynamically allocated, or you're risking crash. Some platforms can have this limit even lower.
When the memory must live after the function returns. Stack memory gets destroyed when function ends, dynamic memory is freed when you want.
When you're building a structure (like array, or graph) of size that is unknown (i.e. may get big), dynamically changes or is too hard to precalculate. Dynamic allocation allows your code to naturally request memory piece by piece at any moment and only when you need it. It is not possible to repeatedly request more and more stack space in a for loop.
Prefer stack allocation otherwise. It is faster and can not leak.
You use dynamic memory when the size of your allocation is not known in advance only on runtime.
For example you ask a user to input names (lets say up to 10 names) and store them in a string array. Since you do not know how much names the user will provide (only on runtime that is) you will have to allocate the array only after you know how much to allocate so you will use dynamic allocation.
You can of course use an array of fixed sized 10 but for larger amounts this will be wasteful
Use dynamic memory allocation, if you don't know exactly how much memory your program will need to allocate at compile-time.
int a[n] for example will limit your array size to n. Also, it allocated n x 4 bytes of memory whether you use it or not. This is allocated on the stack, and the variable n must be known at compile time.
int *a = (int *)malloc(n * sizeof (int)) on the other hand allocated at runtime, on the heap, and the n needs to be known only at runtime, not necessarily at compile-time.
This also ensures you allocate exactly as much memory as you really need. However, as you allocated it at runtime, the cleaning up has to be done by you using free.
You should use dynamic memory when:
If you want your object to persist beyond the scope in which it was created.
Usually, stack sizes are limited and hence if your object occupies a lot of memory then you might run out of stack space in such cases one would usually go for dynamic memory allocation.
Note that c99 standard introduces Variable Length Arrays(VLA) in C so you need not use dynamic memory allocation just because you do not know the array dimensions before hand(unless ofcourse #2 mentioned above is the case)
It is best to avoid dynamic memory allocations as much as you can because it means explicitly managing the memory instead of the automatic mechanism provided by the language.Explicit memory management means that you are prone to make more errors, which might lead to catastrophic effects.
Having said that dynamic memory allocations cannot be avoided always and must be used when the use is imperative(two cases mentioned above).
If you can program without dynamic allocation don't use it!
But a day you will be blocked, and the only way to unblock you will be to use dynamic allocation then now you can use it
Als made an interesting point that you should allocate memory from the heap if your object needs to persist beyond the scope in which it was created. In the code above, you don't need to allocate memory from heap at all. You can rewrite it like this:
#include <stdio.h>
int a = 17;
int main(void)
{
int b = 18; //automatic stack memory
int c[1]; // allocating stack memory. sizeof(int) * 1
c[0] = 19;
printf("a = %d at address %x\n", a, &a);
printf("b = %d at address %x\n", b, &b);
printf("c = %d at address %x\n", c[0], c);
system("PAUSE");
return 0;
}
In fact, as part of the C99 standard (Variable-length array), you can use the [] operator to allocate dynamic space for an array on the stack just as you would normally do to create an array. You don't even need to know the size of the array at compilation time. The compiler will just adjust the esp register (for x86 machines) based on the requested allocation space and you're good to go.
This is actually a much more concise, much more clear question than the one I had asked here before(for any who cares): C Language: Why does malloc() return a pointer, and not the value? (Sorry for those who initially think I'm spamming... I hope it's not construed as the same question since I think the way I phrased it there made it unintentionally misleading)
-> Basically what I'm trying to ask is: Why does a C programmer need a pointer to a dynamically-allocated variable/object? (whatever the difference is between variable/object...)
If a C programmer has the option of creating just 'int x' or just 'int *x' (both statically allocated), then why can't he also have the option to JUST initialize his dynamically-allocated variable/object as a variable (and NOT returning a pointer through malloc())?
*If there are some obscure ways to do what I explained above, then, well, why does malloc() seem the way that most textbooks go about dynamic-allocation?
Note: in the following, byte refers to sizeof(char)
Well, for one, malloc returns a void *. It simply can't return a value: that wouldn't be feasible with C's lack of generics. In C, the compiler must know the size of every object at compile time; since the size of the memory being allocated will not be known until run time, then a type that could represent any value must be returned. Since void * can represent any pointer, it is the best choice.
malloc also cannot initialize the block: it has no knowledge of what's being allocated. This is in contrast with C++'s operator new, which does both the allocation and the initialization, as well as being type safe (it still returns a pointer instead of a reference, probably for historical reasons).
Also, malloc allocates a block of memory of a specific size, then returns a pointer to that memory (that's what malloc stands for: memory allocation). You're getting a pointer because that's what you get: an unitialized block of raw memory. When you do, say, malloc(sizeof(int)), you're not creating a int, you're allocating sizeof(int) bytes and getting the address of those bytes. You can then decide to use that block as an int, but you could also technically use that as an array of sizeof(int) chars.
The various alternatives (calloc, realloc) work roughly the same way (calloc is easier to use when dealing with arrays, and zero-fills the data, while realloc is useful when you need to resize a block of memory).
Suppose you create an integer array in a function and want to return it. Said array is a local variable to the function. You can't return a pointer to a local variable.
However, if you use malloc, you create an object on the heap whose scope exceeds the function body. You can return a pointer to that. You just have to destroy it later or you will have a memory leak.
It's because objects allocated with malloc() don't have names, so the only way to reference that object in code is to use a pointer to it.
When you say int x;, that creates an object with the name x, and it is referenceable through that name. When I want to set x to 10, I can just use x = 10;.
I can also set a pointer variable to point to that object with int *p = &x;, and then I can alternatively set the value of x using *p = 10;. Note that this time we can talk about x without specifically naming it (beyond the point where we acquire the reference to it).
When I say malloc(sizeof(int)), that creates an object that has no name. I cannot directly set the value of that object by name, since it just doesn't have one. However, I can set it by using a pointer variable that points at it, since that method doesn't require naming the object: int *p = malloc(sizeof(int)); followed by *p = 10;.
You might now ask: "So, why can't I tell malloc to give the object a name?" - something like malloc(sizeof(int), "x"). The answer to this is twofold:
Firstly, C just doesn't allow variable names to be introduced at runtime. It's just a basic restriction of the language;
Secondly, given the first restriction the name would have to be fixed at compile-time: if this is the case, C already has syntax that does what you want: int x;.
You are thinking about things wrong. It is not that int x is statically allocated and malloc(sizeof(int)) is dynamic. Both are allocated dynamically. That is, they are both allocated at execution time. There is no space reserved for them at the time you compile. The size may be static in one case and dynamic in the other, but the allocation is always dynamic.
Rather, it is that int x allocates the memory on the stack and malloc(sizeof(int)) allocates the memory on the heap. Memory on the heap requires that you have a pointer in order to access it. Memory on the stack can be referenced directly or with a pointer. Usually you do it directly, but sometimes you want to iterate over it with pointer arithmetic or pass it to a function that needs a pointer to it.
Everything works using pointers. "int x" is just a convenience - someone, somewhere got tired of juggling memory addresses and that's how programming languages with human-readable variable names were born.
Dynamic allocation is... dynamic. You don't have to know how much space you are going to need when the program runs - before the program runs. You choose when to do it and when to undo it. It may fail. It's hard to handle all this using the simple syntax of static allocation.
C was designed with simplicity in mind and compiler simplicity is a part of this. That's why you're exposed to the quirks of the underlying implementations. All systems have storage for statically-sized, local, temporary variables (registers, stack); this is what static allocation uses. Most systems have storage for dynamic, custom-lifetime objects and system calls to manage them; this is what dynamic allocation uses and exposes.
There is a way to do what you're asking and it's called C++. There, "MyInt x = 42;" is a function call or two.
I think your question comes down to this:
If a C programmer has the option of creating just int x or just int *x (both statically allocated)
The first statement allocates memory for an integer. Depending upon the placement of the statement, it might allocate the memory on the stack of a currently executing function or it might allocate memory in the .data or .bss sections of the program (if it is a global variable or static variable, at either file scope or function scope).
The second statement allocates memory for a pointer to an integer -- it hasn't actually allocated memory for the integer itself. If you tried to assign a value using the pointer *x=1, you would either receive a very quick SIGSEGV segmentation violation or corrupt some random piece of memory. C doesn't pre-zero memory allocated on the stack:
$ cat stack.c
#include <stdio.h>
int main(int argc, char *argv[]) {
int i;
int j;
int k;
int *l;
int *m;
int *n;
printf("i: %d\n", i);
printf("j: %d\n", j);
printf("k: %d\n", k);
printf("l: %p\n", l);
printf("m: %p\n", m);
printf("n: %p\n", n);
return 0;
}
$ make stack
cc stack.c -o stack
$ ./stack
i: 0
j: 0
k: 32767
l: 0x400410
m: (nil)
n: 0x4005a0
l and n point to something in memory -- but those values are just garbage, and probably don't belong to the address space of the executable. If we store anything into those pointers, the program would probably die. It might corrupt unrelated structures, though, if they are mapped into the program's address space.
m at least is a NULL pointer -- if you tried to write to it, the program would certainly die on modern hardware.
None of those three pointers actually point to an integer yet. The memory for those integers doesn't exist. The memory for the pointers does exist -- and is initially filled with garbage values, in this case.
The Wikipedia article on L-values -- mostly too obtuse to fully recommend -- makes one point that represented a pretty significant hurdle for me when I was first learning C: In languages with assignable variables it becomes necessary to distinguish between the R-value (or contents) and the L-value (or location) of a variable.
For example, you can write:
int a;
a = 3;
This stores the integer value 3 into whatever memory was allocated to store the contents of variable a.
If you later write:
int b;
b = a;
This takes the value stored in the memory referenced by a and stores it into the memory location allocated for b.
The same operations with pointers might look like this:
int *ap;
ap=malloc(sizeof int);
*ap=3;
The first ap= assignment stores a memory location into the ap pointer. Now ap actually points at some memory. The second assignment, *ap=, stores a value into that memory location. It doesn't update the ap pointer at all; it reads the value stored in the variable named ap to find the memory location for the assignment.
When you later use the pointer, you can choose which of the two values associated with the pointer to use: either the actual contents of the pointer or the value pointed to by the pointer:
int *bp;
bp = ap; /* bp points to the same memory cell as ap */
int *bp;
bp = malloc(sizeof int);
*bp = *ap; /* bp points to new memory and we copy
the value pointed to by ap into the
memory pointed to by bp */
I found assembly far easier than C for years because I found the difference between foo = malloc(); and *foo = value; confusing. I hope I found what was confusing you and seriously hope I didn't make it worse.
Perhaps you misunderstand the difference between declaring 'int x' and 'int *x'. The first allocates storage for an int value; the second doesn't - it just allocates storage for the pointer.
If you were to "dynamically allocate" a variable, there would be no point in the dynamic allocation anyway (unless you then took its address, which would of course yield a pointer) - you may as well declare it statically. Think about how the code would look - why would you bother with:
int x = malloc(sizeof(int)); *x = 0;
When you can just do:
int x = 0;