I've been trying to get into the habit of defining trivial variables at the point they're needed. I've been cautious about writing code like this:
while (n < 10000) {
int x = foo();
[...]
}
I know that the standard is absolutely clear that x exists only inside the loop, but does this technically mean that the integer will be allocated and deallocated on the stack with every iteration? I realise that an optimising compiler isn't likely to do this, but it that guaranteed?
For example, is it ever better to write:
int x;
while (n < 10000) {
x = foo();
[...]
}
I don't mean with this code specifically, but in any kind of loop like this.
I did a quick test with gcc 4.7.2 for a simple loop differing in this way and the same assembly was produced, but my question is really are these two, according to the standard, identical?
Note that "allocating" automatic variables like this is pretty much free; on most machines it's either a single-instruction stack pointer adjustment, or the compiler uses registers in which case nothing needs to be done.
Also, since the variable remains in scope until the loop exits, there's absolutely no reason to "delete" (=readjust the stack pointer) it until the loop exits, I certainly wouldn't expect there to be any overhead per-iteration for code like this.
Also, of course the compiler is free to "move" the allocation out of the loop altogether if it feels like it, making the code equivalent to your second example with the int x; before the while. The important thing is that the first version is easier to read and more tighly localized, i.e. better for humans.
Yes, the variable x inside the loop is technically defined on each iteration, and initialized via the call to foo() on each iteration. If foo() produces a different answer each time, this is fine; if it produces the same answer each time, it is an optimization opportunity – move the initialization out of the loop. For a simple variable like this, the compiler typically just reserves sizeof(int) bytes on the stack — if it can't keep x in a register — that it uses for x when x is in scope, and may reuse that space for other variables elsewhere in the same function. If the variable was a VLA — variable length array — then the allocation is more complex.
The two fragments in isolation are equivalent, but the difference is the scope of x. In the example with x declared outside the loop, the value persists after the loop exits. With x declared inside the loop, it is inaccessible once the loop exits. If you wrote:
{
int x;
while (n < 10000)
{
x = foo();
...other stuff...
}
}
then the two fragments are near enough equivalent. At the assembler level, you'll be hard pressed to spot the difference in either case.
My personal point of view is that once you start worrying about such micro-optimisations, you're doomed to failure. The gain is:
a) Likely to be very small
b) Non-portable
I'd stick with code that makes your intention clear (i.e. declare x inside the loop) and let the compiler care about efficiency.
There is nothing in the C standard that says how the compiler should generate code in either case. It could adjust the stack pointer on every iteration of the loop if it fancies.
That being said, unless you start doing something crazy with VLAs like this:
void bar(char *, char *);
void
foo(int x)
{
int i;
for (i = 0; i < x; i++) {
char a[i], b[x - i];
bar(a, b);
}
}
the compiler will most likely just allocate one big stack frame at the beginning of the function. It's harder to generate code for creating and destroying variables in blocks instead of just allocating all you need at the beginning of the function.
Related
#include <stdio.h>
int main()
{
for(int i=0;i<100;i++)
{
int count=0;
printf("%d ",++count);
}
return 0;
}
output of the above program is: 1 1 1 1 1 1..........1
Please take a look at the code above. I declared variable "int count=0" inside the for loop.
With my knowledge, the scope of the variable is within the block, so count variable will be alive up to for loop execution.
"int count=0" is executing 100 times, then it has to create the variable 100 times else it has to give the error (re-declaration of the count variable), but it's not happening like that — what may be the reason?
According to output the variable is initializing with zero every time.
Please help me to find the reason.
Such simple code can be visualised on http://www.pythontutor.com/c.html for easy understanding.
To answer your question, count gets destroyed when it goes outside its scope, that is the closing } of the loop. On next iteration, a variable of the same name is created and initialised to 0, which is used by the printf.
And if counting is your goal, print i instead of count.
The C standard describes the C language using an abstract model of a computer. In this model, count is created each time the body of the loop is executed, and it is destroyed when execution of the body ends. By “created” and “destroyed,” we mean that memory is reserved for it and is released, and that the initialization is performed with the reservation.
The C standard does not require compilers to implement this model slavishly. Most compilers will allocate a fixed amount of stack space when the routine starts, with space for count included in this fixed amount, and then count will use that same space in each iteration. Then, if we look at the assembly code generated, we will not see any reservation or release of memory; the stack will be grown and shrunk only once for the whole routine, not grown and shrunk in each loop iteration.
Thus, the answer is twofold:
In C’s abstract model of computing, a new lifetime of count begins and ends in each loop iteration.
In most actual implementations, memory is reserved just once for count, although implementations may also allocate and release memory in each iteration.
However, even if you know your C implementation allocates stack space just once per routine when it can, you should generally think about programs in the C model in this regard. Consider this:
for (int i = 0; i < 100; ++i)
{
int count = 0;
// Do some things with count.
float x = 0;
// Do some things with x.
}
In this code, the compiler might allocate four bytes of stack space to use for both count and x, to be used for one of them at a time. The routine would grow the stack once, when it starts, including four bytes to use for count and x. In each iteration of the loop, it would use the memory first for count and then for x. This lets us see that the memory is first reserved for count, then released, then reserved for x, then released, and then that repeats in each iteration. The reservations and releases occur conceptually even though there are no instructions to grow and shrink the stack.
Another illuminating example is:
for (int i = 0; i < 100; ++i)
{
extern int baz(void);
int a[baz()], b[baz()];
extern void bar(void *, void *);
bar(a, b);
}
In this case, the compiler cannot reserve memory for a and b when the routine starts because it does not know how much memory it will need. In each iteration, it must call baz to find how much memory is needed for a and how much for b, and then it must allocate stack space (or other memory) for them. Further, since the sizes may vary from iteration to iteration, it is not possible for both a and b to start in the same place in each iteration—one of them must move to make way for the other. So this code lets us see that a new a and a new b must be created in each iteration.
int count=0 is executing 100 times, then it has to create the variable 100 times
No, it defines the variable count once, then assigns it the value 0 100 times.
Defining a variable in C does not involve any particular step or code to "create" it (unlike for example in C++, where simply defining a variable may default-construct it). Variable definitions just associate the name with an "entity" that represents the variable internally, and definitions are tied to the scope where they appear.
Assigning a variable is a statement which gets executed during the normal program flow. It usually has "observable effects", otherwise the compiler is allowed to optimize it out entirely.
OP's example can be rewritten in a completely equivalent form as follows.
for(int i=0;i<100;i++)
{
int count; // definition of variable count - defined once in this {} scope
count=0; // assignment of value 0 to count - executed once per iteration, 100 times total
printf("%d ",++count);
}
Eric has it correct. In much shorter form:
Typically compilers determine at compile time how much memory is needed by a function and the offsets in the stack to those variables. The actual memory allocations occur on each function call and memory release on the function return.
Further, when you have variables nested within {curly braces} once execution leaves that brace set the compiler is free to reuse that memory for other variables in the function. There are two reasons I intentionally do this:
The variables are large but only needed for a short time so why make stacks larger than needed? Especially if you need several large temporary structures or arrays at different times. The smaller the scope the less chance of bugs.
If a variable only has a sane value for a limited amount of time, and would be dangerous or buggy to use out of that scope, add extra curly braces to limit the scope of access so improper use generates immediate compiler errors. Using unique names for each variable, even if the compiler doesn't insist on it, can help the debugger, and your mind, less confused.
Example:
your_function(int a)
{
{ // limit scope of stack_1
int stack_1 = 0;
for ( int ii = 0; ii < a; ++ii ) { // really limit scope of ii
stack_1 += some_calculation(i, a);
}
printf("ii=%d\n", ii); // scope error
printf("stack_1=%d\n", stack_1); // good
} // done with stack_1
{
int limited_scope_1[10000];
do_something(a,limited_scope_1);
}
{
float limited_scope_2[10000];
do_something_else(a,limited_scope_2);
}
}
A compiler given code like:
void do_something(int, int*);
...
for (int i=0; i<100; i++)
{
int const j=(i & 1);
doSomething(i, &j);
}
could legitimately replace it with:
void do_something(int, int*);
...
int const __compiler_generated_0 = 0;
int const __compiler_generated_1 = 1;
for (int i=0; i<100; i+=2)
{
doSomething(i, &compiler_generated_0);
doSomething(i+1, &compiler_generated_1);
}
Although a compiler would typically allocate space on the stack once for j, when the function was entered, and then not reuse the storage during the loop (or even the function), meaning that j would have the same address on every iteration of the loop, there is no requirement that the address remain constant. While there typically wouldn't be an advantage to having the address vary on different iterations, compilers are be allowed to exploit such situations should they arise.
Depended of the version of the C compiler and compiler flags it is possible to initialize variables on any place in your functions (As far as I am aware).
I'm used to put it all the variables at the top of the function, but the discussion started about the memory use of the variables if defined in any other place in the function.
Below I have written 2 short examples, and I wondered if anyone could explain me (or verify) how the memory gets allocated.
Example 1: Variable y is defined after a possible return statement, there is a possibility this variable won't be used for that reason, as far as I'm aware this doesn't matter and the code (memory allocation) would be the same if the variable was placed at the top of the function. Is this correct?
Example 2: Variable x is initialized in a loop, meaning that the scope of this variable is only within this loop, but what about the memory use of this variable? Would it be any different if placed on the top of the functions? Or just initialized on the stack at the function call?
Edit: To conclude a main question:
Does reducing the scope of the variable or change the location of the first use (so anywhere else instead of top) have any effects on the memory use?
Code example 1
static void Function(void){
uint8_t x = 0;
//code changing x
if(x == 2)
{
return;
}
uint8_t y = 0;
//more code changing y
}
Code example 2
static void LoopFunction(void){
uint8_t i = 0;
for(i =0; i < 100; i ++)
{
uint8_t x = i;
// do some calculations
uartTxLine("%d", x);
}
//more code
}
I'm used to put it all the variables at the top of the function
This used to be required in the older versions of C, but modern compilers dropped that requirement. As long as they know the type of the variable at the point of its first use, the compilers have all the information they need.
I wondered if anyone could explain me how the memory gets allocated.
The compiler decides how to allocate memory in the automatic storage area. Implementations are not limited to the approach that gives each variable you declare a separate location. They are allowed to reuse locations of variables that go out of scope, and also of variables no longer used after a certain point.
In your first example, variable y is allowed to use the space formerly occupied by variable x, because the first point of use of y is after the last point of use of x.
In your second example the space used for x inside the loop can be reused for other variables that you may declare in the // more code area.
Basically, the story goes like this. When calling a function in raw assembler, it is custom to store everything used by the function on the stack upon entering the function, and clean it up upon leaving. Certain CPUs and ABIs may have a calling convention which involves automatic stacking of parameters.
Likely because of this, C and many other old languages had the requirement that all variables must be declared at the top of the function (or on top of the scope), so that the { } reflect push/pop on the stack.
Somewhere around the 80s/90s, compilers started to optimize such code efficiently, as in they would only allocate room for a local variable at the point where it was first used, and de-allocate it when there was no further use for it. Regardless of where that variable was declared - it didn't matter for the optimizing compiler.
Around the same time, C++ lifted the variable declaration restrictions that C had, and allowed variables to be declared anywhere. However, C did not actually fix this before the year 1999 with the updated C99 standard. In modern C you can declare variables everywhere.
So there is absolutely no performance difference between your two examples, unless you are using an incredibly ancient compiler. It is however considered good programming practice to narrow the scope of a variable as much as possible - though it shouldn't be done at the expense of readability.
Although it is only a matter of style, I would personally prefer to write your function like this:
(note that you are using the wrong printf format specifier for uint8_t)
#include <inttypes.h>
static void LoopFunction (void)
{
for(uint8_t i=0; i < 100; i++)
{
uint8_t x = i;
// do some calculations
uartTxLine("%" PRIu8, x);
}
//more code
}
Old C allowed only to declare (and initialize) variables at the top of a block. You where allowed to init a new block (a pair of { and } characters) anywhere inside a block, so you had then the possibility of declaring variables next to the code using them:
... /* inside a block */
{ int x = 3;
/* use x */
} /* x is not adressabel past this point */
And you where permitted to do this in switch statements, if statements and while and do statements (everywhere where you can init a new block)
Now, you are permitted to declare a variable anywhere where a statement is allowed, and the scope of that variable goes from the point of declaration to the end of the inner nested block you have declared it into.
Compilers decide when they allocate storage for local variables so, you can all of them allocated when you create a stack frame (this is the gcc way, as it allocates local variables only once) or when you enter in the block of definition (for example, Microsoft C does this way) Allocating space at runtime is something that requires advancing the stack pointer at runtime, so if you do this only once per stack frame you are saving cpu cycles (but wasting memory locations). The important thing here is that you are not allowed to refer to a variable location outside of its scoping definition, so if you try to do, you'll get undefined behaviour. I discovered an old bug for a long time running over internet, because nobody take the time to compile that program using Microsoft-C compiler (which failed in a core dump) instead of the commmon use of compiling it with GCC. The code was using a local variable defined in an inner scope (the then part of an if statement) by reference in some other part of the code (as everything was on main function, the stack frame was present all the time) Microsoft-C just reallocated the space on exiting the if statement, but GCC waited to do it until main finished. The bug solved by just adding a static modifier to the variable declaration (making it global) and no more refactoring was neccesary.
int main()
{
struct bla_bla *pointer_to_x;
...
if (something) {
struct bla_bla x;
...
pointer_to_x = &x;
}
/* x does not exist (but it did in gcc) */
do_something_to_bla_bla(pointer_to_x); /* wrong, x doesn't exist */
} /* main */
when changed to:
int main()
{
struct bla_bla *pointer_to_x;
...
if (something) {
static struct bla_bla x; /* now global ---even if scoped */
...
pointer_to_x = &x;
}
/* x is not visible, but exists, so pointer_to_x continues to be valid */
do_something_to_bla_bla(pointer_to_x); /* correct now */
} /* main */
Can anyone tell me the effect that the below code has on memory?
My questions are:
In code 1, is the same memory location getting updated every
time in the loop?
In code 2, is new memory allocated as the variable is declared and assigned in for loop?
Code 1:
int num;long result;
for(int i=0;i<500;i++){
num = i;
result = num * num;
}
Code 2:
for(int i=0;i<500;i++){
int num = i;
long result = num * num;
}
In both cases only one num and one result instance will be created.
The only different is that on Code 2 num and result won't be accessible after the loop and the memory used to hold them can be reused for other members.
Important: Where you declare a local variable in your source code has very little impact on when the actual allocation and deallocation (for example push/pop on the stack) takes place. If at all, the variable might get optimized away entirely.
Where in your C code you allocate your local variable has most likely no impact on performance what-so-ever. Therefore, you should declare local variables where it gives the best possible readability.
In both cases, the compiler will deduce that num is completely superfluous and optimize it away. No memory will be allocated for it.
In code 2, the compiler will deduce that the local variable result isn't used anywhere in the program, and therefore it will most likely optimize away the whole of code 2 into nothing.
The machine code of code 1 will look something like this:
allocate room for i
allocate room for result
set i to 0
loop:
multiply i by i
store in result
if i is less than 500, jump to loop
Can anyone tell me the effect that the below code has on memory?
As others have mentioned, the answer to your question hugely depends on the compiler you are using.
[1] Assuming you intend to use result and num later in the loop for other computations as in:
for(int i=0; i<500; ++i){
int num = i;
long result = num * num;
// more operations using num and result, e.g. function calls, IO, etc.
}
, a modern compiler (gcc, llvm, mvc, icc) would be smart enough to optimise the two codes you provided to the same thing, thus in both codes the same "memory location" would be updated on each iteration.
I put "memory location" in quotes, as the variables can be promoted to registers, which, while strictly speaking are still memory, are a lot faster to access and thus a preferable location for frequently used variables.
The only difference between your codes is the scope of your variables, but you probably already know that.
If, conversly to [1], you don't intend to use your variables later, the compiler would probably detect that and just skip the loop, not generating any machine code for it, as it is redundant.
If I define an array in if statement then does memory gets allocated during compile time eg.
if(1)
{
int a[1000];
}
else
{
float b[1000];
}
Then a memory of 2 * 1000 for ints + 4 * 1000 for floats get allocated?
It is reserved on the stack at run-time (assuming a non-trivial condition - in your case, the compiler would just exclude the else part). That means it only exists inside the scope block (between the {}).
In your example, only the memory for the ints gets allocated on the stack (1000 * sizeof(int)).
As you can guess, this is happening at run time. The generated code has instructions to allocate the space on the stack when the corresponding block of code is entered.
Keep in mind that this is happening because of the semantics of the language. The block structure introduces a new scope, and any automatic variables allocated in that scope have a lifetime that lasts as long as the scope does. In C, this is implemented by allocating it on the stack, which collapses as the scope disappears.
Just to drive home the point, note that the allocation would be different had the variables been of different nature.
if(1)
{
static int a[1000];
}
else
{
static float b[1000];
}
In this case, space is allocated for both the ints and the floats. The lifetime of these variables is the program. But the visibility is within the block scope they are allocated in.
Scope
Variables declared inside the scope of a pair of { } are on the stack. This applies to variables declared at the beginning of a function or in any pair of { } within the function.
int myfunc()
{
int i = 0; // On the stack, scoped: myfunc
printf("%i\n");
if (1)
{
int j = 1; // On the stack, scope: this if statement
printf("%i %i\n",i,j);
}
printf("%i %i\n",i,j); // Won't work, no j
}
These days the scope of the variables is limited to the surrounding { }. I recall that some older Microsoft compilers didn't limit the scope, and that in the example above the final printf() would compile.
So Where is it in memory?
The memory of i and j is merely reserved on the stack. This is not the same as memory allocation done with malloc(). That is important, because calling malloc() is very slow in comparison. Also with memory dynamically allocated using malloc() you have to call free().
In effect the compiler knows ahead of time what space is needed for a function's variables and will generate code that refers to memory relative to whatever the stack pointer is when myfunc() is called. So long as the stack is big enough (2MBytes normally, depends on the OS), all is good.
Stack overflow occurs in the situation where myfunc() is called with the stack pointer already close to the end of the stack (i.e. myfunc() is called by a function which in turn had been called by another which it self was called by yet another, etc. Each layer of nested calls to functions moves the stack pointer on a bit more, and is only moved back when functions return).
If the space between the stack pointer and the end of the stack isn't big enough to hold all the variables that are declared in myfunc(), the code for myfunc() will simply try to use locations beyond the end of the stack. That is almost always a bad thing, and exactly how bad and how hard it is to notice that something has gone wrong depends on the operating system. On small embedded micro controllers it can be a nightmare as it usually means some other part of the program's data (eg global variables) get silently overwritten, and it can be very hard to debug. On bigger systems (Linux, Windows) the OS will tell you what's happened, or will merely make the stack bigger.
Runtime Efficiency Considerations
In the example above I'm assigning values to i and j. This does actually take up a small amount of runtime. j is assigned 1 only after evaluation of the if statement and subsequent branch into where j is declared.
Say for example the if statement hadn't evaluated as true; in that case j is never assigned 1. If j was declared at the start of myfunc() then it would always get assigned the value of 1 regardless of whether the if statement was true - a minor waste of time. But consider a less trivial example where a large array is declared an initialised; that would take more execution time.
int myfunc()
{
int i = 0; // On the stack, scoped: myfunc
int k[10000] = {0} // On the stack, scoped: myfunc. A complete waste of time
// when the if statement evaluates to false.
printf("%i\n");
if (0)
{
int j = 1; // On the stack, scope: this if statement
// It would be better to move the declaration of k to here
// so that it is initialised only when the if evaluates to true.
printf("%i %i %i\n",i,j,k[500]);
}
printf("%i %i\n",i,j); // Won't work, no j
}
Placing the declaration of k at the top of myfunc() means that a loop 10,000 long is executed to initialise k every time myfunc() is called. However it never gets used, so that loop is a complete waste of time.
Of course, in these trivial examples compilers will optimise out the unnecessary code, etc. In real code where the compiler cannot predict ahead of time what the execution flow will be then things are left in place.
Memory for the array in the if block will be allocated on stack at run time. else part will be optimized (removed) by the compiler. For more on where the variables will be allocated memory, see Segmentation Fault when writing to a string
As DCoder & paddy corrected me, the memory will be calculated at compile time but allocated at run-time in stack memory segment, but with the scope & lifetime of the block in which the array is defined. The size of memory allocated depends on size of int & float in your system. Read this for an overview on C memory map
I know that sometimes if you don't initialize an int, you will get a random number if you print the integer.
But initializing everything to zero seems kind of silly.
I ask because I'm commenting up my C project and I'm pretty straight on the indenting and it compiles fully (90/90 thank you Stackoverflow) but I want to get 10/10 on the style points.
So, the question: when is it appropriate to initialize, and when should you just declare a variable:
int a = 0;
vs.
int a;
There are several circumstances where you should not initialize a variable:
When it has static storage duration (static keyword or global var) and you want the initial value to be zero. Most compilers will actually store zeros in the binary if you explicitly initialize, which is usually just a waste of space (possibly a huge waste for large arrays).
When you will be immediately passing the address of the variable to another function that fills its value. Here, initializing is just a waste of time and may be confusing to readers of the code who wonder why you're storing something in a variable that's about to be overwritten.
When a meaningful value for the variable can't be determined until subsequent code has completed execution. In this case, it's actively harmful to initialize the variable with a dummy value such as zero/NULL, as this prevents the compiler from warning you if you have some code paths where a meaningful value is never assigned. Compilers are good at warning you about accessing uninitialized variables, but can't warn you about "still contains dummy value" variables.
Aside from these issues, I'd say it's generally good practice to initialize your non-static variables when possible.
A rule that hasn't been mentioned yet is this: when the variable is declared inside a function it is not initialised, and when it is declared in static or global scope it's set to 0:
int a; // is set to 0
void foo() {
int b; // set to whatever happens to be in memory there
}
However - for readability I would usually initialise everything at declaration time.
If you're interested in learning this sort of thing in detail, I'd recommend this presentation and this book
I can think of a couple of reason off the top of my head:
When you're going to be initializing it later on in your code.
int x;
if(condition)
{
func();
x = 2;
}
else
{
x = 3;
}
anotherFunc(x); // x will have been set a value no matter what
When you need some memory to store a value set by a function or another piece of code:
int x; // Would be pointless to give x a value here
scanf("%d", &x);
If the variable is in the scope of of a function and not a member of a class I always initialize it because otherwise you will get warnings. Even if this variable will be used later I prefer to assign it on declaration.
As for member variables, you should initialize them in the constructor of your class.
For pointers, always initialize them to some default, particularly NULL, even if they are to be used later, they are dangerous when uninitialized.
Also it is recommended to build your code with the highest level of warnings that your compiler supports, it helps to identify bad practices and potential errors.
Static and global variables will be initialized to zero for you so you may skip initialization. Automatic variables (e.g. non-static variables defined in function body) may contain garbage and should probably always be initialized.
If there is a non-zero specific value you need at initialization then you should always initialize explicitly.
It's always good practice to initialize your variables, but sometimes it's not strictly necessary. Consider the following:
int a;
for (a = 0; a < 10; a++) { } // a is initialized later
or
void myfunc(int& num) {
num = 10;
}
int a;
myfunc(&a); // myfunc sets, but does not read, the value in a
or
char a;
cin >> a; // perhaps the most common example in code of where
// initialization isn't strictly necessary
These are just a couple of examples where it isn't strictly necessary to initialize a variable, since it's set later (but not accessed between declaration and initialization).
In general though, it doesn't hurt to always initialize your variables at declaration (and indeed, this is probably best practice).
In general, there's no need to initialize a variable, with 2 notable exceptions:
You're declaring a pointer (and not assigning it immediately) - you
should always set these to NULL as good style and defensive
programming.
If, when you declare the variable, you already know
what value is going to be assigned to it. Further assignments use up
more CPU cycles.
Beyond that, it's about getting the variables into the right state that you want them in for the operation you're going to perform. If you're not going to be reading them before an operation changes their value (and the operation doesn't care what state it is in), there's no need to initialize them.
Personally, I always like to initialize them anyway; if you forgot to assign it a value, and it's passed into a function by mistake (like a remaining buffer length) 0 is usually cleanly handled - 32532556 wouldn't be.
There is absolutely no reason why variables shouldn't be initialised, the compiler is clever enough to ignore the first assignment if a variable is being assigned twice. It is easy for code to grow in size where things you took for granted (such as assigning a variable before being used) are no longer true. Consider:
int MyVariable;
void Simplistic(int aArg){
MyVariable=aArg;
}
//Months later:
int MyVariable;
void Simplistic(int aArg){
MyVariable+=aArg; // Unsafe, since MyVariable was never initialized.
}
One is fine, the other lands you in a heap of trouble. Occasionally you'll have issues where your application will run in debug mode, but release mode will throw an exception, one reason for this is using an uninitialised variable.
As long as I have not read from a variable before writing to it, I have not had to bother with initializing it.
Reading before writing can cause serious and hard to catch bugs. I think this class of bugs is notorious enough to gain a mention in the popular SICP lecture videos.
Initializing a variable, even if it is not strictly required, is ALWAYS a good practice. The few extra characters (like "= 0") typed during development may save hours of debugging time later, particularly when it is forgotten that some variables remained uninitialized.
In passing, I feel it is good to declare a variable close to its use.
The following is bad:
int a; // line 30
...
a = 0; // line 40
The following is good:
int a = 0; // line 40
Also, if the variable is to be overwritten right after initialization, like
int a = 0;
a = foo();
it is better to write it as
int a = foo();