I'm trying to understand a coding example I found in C, and one of the for loops is structured like this:
for(int x=0; int y=0; int z!=0){some code}
I am used to seeing something more like this:
for(int x=0; x < someLimit; x++){some code}
What does the mystery for loop do exactly?
You copied the statement wrong. In the program you cited in your comment there's only one for statement which approximates what you posted above:
for(time=0,count=0;remain!=0;)
In this case the "initialization" portion of the for statement is
time=0,count=0
Note that the character between the initialization of time and count is a comma, not a semicolon. This means
that both time and count are set to zero. The "test" portion of the for statement is
remain != 0
meaning that the loop will continue as long as remain is not equal to zero.
In this for statement the "increment" portion is empty - so nothing is incremented/decremented/whatever at the end of each pass through the loop.
Best of luck.
In for loop you can omit initialization part(provided you initialized your variable before) and increment/decrement part. you can have also multiple initialization like in your example x=0;y=0 and multiple increment/decrement. in your first example for loop have multiple initialization for x and y and no increment/decrement part. and z part is the condition. Please check why you are declaring z there. I am not sure about that. It would be better if you will upload full code of that example.
Related
I am a beginner is Computer Science and I recently started learning the language C.
I was studying the for loop and in the book it was written that even if we replace the initialization;testing;incrementation statement of a for loop by any valid statement the compiler will not show any syntax error.
So now I run the following program:
#include<stdio.h>
int main()
{
int i;
int j;
for(i<4;j=5;j=0)
printf("%d",i);
return 0;
}
I have got the following output.
OUTPUT:
1616161616161616161616161616161616161616.........indefinitely
I understood why this is an indefinite loop but i am unable to understand why my PC is printing this specific output? Is there any way to understand in these above kind of programs what the system will provide us as output?
This is a case of undefined behaviour.
You have declared i so it has a memory address but haven't set the value so its value is just whatever was already in that memory address (in this case 16)
I'd guess if you ran the code multiple times (maybe with restarts between) the outputs would change.
Some more information on uninitialized variables and undefined behaviour: https://www.learncpp.com/cpp-tutorial/uninitialized-variables-and-undefined-behavior/
Looks like i was never initialized, and happens to contain 16, which the for loop is printing continuously. Note that printf does not add a new line automatically. Probably want to read the manual on how to use for.
To better understand what is going on, you can add additional debugging output to see what other variables are set to (don't forget the \n newline!) but really problem is the for loop doesn't seem right. Unless there's something going on with j that you aren't showing us, it really doesn't belong there at all.
You are defining the variable i but never initializing it, instead, you are defining the value for j but never using it.
On a for loop, first you initialize the control variable, if you haven't already done it. Then you specify the condition you use to know if you should run another iteration or not. And last but not least, change the value of the control variable so you won't have infinite loops.
An example:
#include <stdio.h>
int main()
{
// 1. we declare i as an integer starting from 1
// 2. we will keep iterating as long as i is smaller than 10
// 3. at the end of each iteration, we increment it's value by 1
for (int i = 1; i < 10; i++) {
printf("%d\n", i);
}
return 0;
}
so I'm slowly trying to learn C from scratch.
I'm at a point in the book I'm using where an exercise is proposed: use nested loops to find Pythagorean triplets. Now I'll show the code.
#include <stdio.h>
int main(void){
int lato1=1;
int lato2=1;
int ipotenusa=1;
for(;lato1 <= 500; lato1++){
for(;lato2 <= 500; lato2++){
for(;ipotenusa <= 500; ipotenusa++){
if (((lato1 * lato1)+(lato2 * lato2))==(ipotenusa*ipotenusa)){
printf("Tripletta %10d %10d %10d\n",lato1,lato2,ipotenusa);
}
}
}
}
return 0;
}
now, apart from terrible formatting and style, apart from the shi*t optimization,
the code as shown, doesnt work.
It only execute the inner most loop, and then the program ends.
If, however, I initialize each variable/counter inside each loop then it works.
Why?
I read that for loop initialization is valid even whitout no arguments (;;) but in this case I just wanted to initialize those variables before the loop, let's say because I wanted to access those value after the loop is done, it just doesn't seem to work like it's supposed to.
English is not my primary language so apologies in advance for any mystake.
Can somebody explain what is the problem?
Edit 1: So, I don't know if it was my bad English or something else.
I said, if I declare and initialize the variables before the loop, like in the code I've shown you, it only goes through the inner most loop (ipotenusa) and it does so with the following output values: 1 1 1 then 1 1 2 then 1 1 3 and so on, where the only increasing number is the last one (ipotenusa); AFTER we reach 1 1 500 the programs abruptily ends.
I then said that if I initialize the variables as normal, meaning inside the for loop instruction, then it works as intended.
Even if declared earlier there is NO reason it should't work. The variable's value it's supposed to increase. Up until now the only useful answer was to initialize a variable outside the loop but assign a value to it in the loop statement, but this is not at the end the answer I need, because I should be able to skip the initialization inside the loop statement altogether.
EDIT 2: I was wrong and You guys were right, language barrier (most likely foolishness, rather) was certainly the cause of the misunderstanding lol. Sorry and Thanks for the help!
You accidentally answered your own question:
...in this case I just wanted to initialize those variables before the loop, let's say because I wanted to access those value after the loop is done...
Think about the value of ipotenusa...or have your program print out the value of each variable at the start of each loop for debugging purposes. It'll look something like the following log.
lato1: 1
lato2: 1
ipotenusa: 1
ipotenusa: 2
ipotenusa: 3
ipotenusa: 4
ipotenusa: 5
lato2: 2
lato2: 3
lato2: 4
lato2: 5
lato1: 2
lato1: 3
lato1: 4
lato1: 5
The inner loops never "reset," because you're saving the value. You're not wrong on either count, but you've set up two requirements that conflict with each other, and the C runtime can only give you one.
Basically, lato2 and ipotenusa need to be initialized (but not necessarily declared) in their for statements, so that they're set up to run again.
int lato1=1;
int lato2;
int ipotenusa;
for(;lato1 <= 5; lato1++){
for(lato2 = 1;lato2 <= 5; lato2++){
for(ipotenusa = 1;ipotenusa <= 5; ipotenusa++){
if (((lato1 * lato1)+(lato2 * lato2))==(ipotenusa*ipotenusa)){
printf("Tripletta %10d %10d %10d\n",lato1,lato2,ipotenusa);
}
}
}
}
It's worth pointing out that, even though the specification might (or might not; I don't remember and am too fuzzy to look it up, right now) say that the loop variables only exist during the loop, I've never seen an implementation actually do that, so you'll see a lot of code out there that defines the variable inside of for and uses it after. It's probably not great and a static checker like lint might complain, but it's common enough that it'll pass most code reviewers, to give you a sense of whether it can be done.
Your code does exactly what you told it to do.
If you initialise lato2 with a value of 1, what makes you think that the initialisation will be repeated at the start of the inner for loop? Of course it isn't. You told the compiler that you didn't want it to be initialised again. After that loop executed the first time, lato2 has a value of 501, and it will keep that value forever.
Even if it hadn't produced a bug, putting the initialisation and the for loop so far apart is very, very bad style (wouldn't pass a code review in a professional setting and therefore would have to be changed).
for (int lato2 = 1; lato2 <= 500; ++lato2) is clear, obvious, and works.
One approach would be - declaring variables outside of the loop and initialize them inside. This way, you can access the variables after the loops.
#include <stdio.h>
int main(void){
int lato1, lato2, ipotenusa;
for(lato1=1; lato1 <= 500; lato1++){
for(lato2=1; lato2 <= 500; lato2++){
for(ipotenusa=1; ipotenusa <= 500; ipotenusa++){
if (((lato1 * lato1)+(lato2 * lato2))==(ipotenusa*ipotenusa)){
printf("Tripletta %10d %10d %10d\n",lato1,lato2,ipotenusa);
}
}
}
}
return 0;
}
You are using variables that are initialized outside of the for loops. Usually what happens is when we use inner loops(nested loops) we intended to initialize it with base value everytime but in your case it is never going to do that. Since you have initialized variable out side of the for loops so it will persist its value through out its life time of course which is main function in this case.
I saw this question in a test in which we have to tell the output of the following code.
#include<stdio.h>
int main(){
int k = 0;
while(+(+k--)!=0)
k=k++;
printf("%d\n", k);
return 0;
}
The output is -1. I am unsure why this is the answer, though.
What does the expression +(+k--) mean in C?
This code is deeply, perhaps deliberately, confusing. It contains a narrowly-averted instance of the dread undefined behavior. It's hard to know whether the person who constructed this question was being very, very clever or very, very stupid. And the "lesson" this code might purport to teach or quiz you about -- namely, that the unary plus operator doesn't do much -- is not one that's important enough, I would think, to deserve this kind of subversive misdirection.
There are two confusing aspects of the code, the strange condition:
while(+(+k--)!=0)
and the demented statement it controls:
k=k++;
I'm going to cover the second part first.
If you have a variable like k that you want to increment by 1, C gives you not one, not two, not three, but four different ways to do it:
k = k + 1
k += 1
++k
k++
Despite this bounty (or perhaps because of it), some programmers get confused and cough out contortions like
k = k++;
If you can't figure out what this is supposed to do, don't worry: no one can. This expression contains two different attempts to alter k's value (the k = part, and the k++ part), and because there's no rule in C to say which of the attempted modifications "wins", an expression like this is formally undefined, meaning not only that it has no defined meaning, but that the whole program containing it is suspect.
Now, if you look very carefully, you'll see that in this particular program, the line k = k++ doesn't actually get executed, because (as we're about to see) the controlling condition is initially false, so the loop runs 0 times. So this particular program might not actually be undefined -- but it's still pathologically confusing.
See also these canonical SO answers to all questions concerning Undefined Behavior of this sort.
But you didn't ask about the k=k++ part. You asked about the first confusing part, the +(+k--)!=0 condition. This looks strange, because it is strange. No one would ever write such code in a real program. So there's not much reason to learn how to understand it. (Yes, it's true, exploring the boundaries of a system can help you learn about its fine points, but there's a line in my book between imaginative, thought-provoking explorations versus dunderheaded, abusive explorations, and this expression is pretty clearly on the wrong side of that line.)
Anyway, let's examine +(+k--)!=0. (And after doing so, let's forget all about it.) Any expression like this has to be understood from the inside out. I presume you know what
k--
does. It takes k's current value and "returns" it to the rest of the expression, and it more or less simultaneously decrements k, that is, it stores the quantity k-1 back into k.
But then what does the + do? This is unary plus, not binary plus. It's just like unary minus. You know that binary minus does subtraction: the expression
a - b
subtracts b from a. And you know that unary minus negates things: the expression
-a
gives you the negative of a. What unary + does is... basically nothing. +a gives you a's value, after changing positive values to positive and negative values to negative. So the expression
+k--
gives you whatever k-- gave you, that is, k's old value.
But we're not done, because we have
+(+k--)
This just takes whatever +k-- gave you, and applies unary + to it again. So it gives you whatever +k-- gave you, which was whatever k-- gave you, which was k's old value.
So in the end, the condition
while(+(+k--)!=0)
does exactly the same thing as the much more ordinary condition
while(k-- != 0)
would have done. (It also does the same thing as the even more complicated-looking condition while(+(+(+(+k--)))!=0) would have done. And those parentheses aren't really necessary; it also does the same thing as while(+ +k--!=0) would have done.)
Even figuring out what the "normal" condition
while(k-- != 0)
does is kind of tricky. There are sort of two things going on in this loop: As the loop runs potentially multiple times, we're going to:
keep doing k--, to make k smaller and smaller, but also
keep doing the body of the loop, whatever that does.
But we do the k-- part right away, before (or in the process of) deciding whether to take another trip through the loop. And remember that k-- "returns" the old value of k, before decrementing it. In this program, the initial value of k is 0. So k-- is going to "return" the old value 0, then update k to -1. But then the rest of the condition is != 0 -- and it's not true that 0 != 0. That is, 0 is equal to 0, so we won't make any trips through the loop, so we won't try to execute the problematic statement k=k++ at all.
In other words, in this particular loop, although I said that "there are sort of two things going on", it turns out that thing 1 happens one time, but thing 2 happens zero times.
At any rate, I hope it's now adequately clear why this poor excuse for a program ends up printing -1 as the final value of k. Normally, I don't like to answer quiz questions like this -- it feels like cheating -- but in this case, since I fundamentally disagree with the whole point of the exercise, I don't mind.
At first glance it looks like this code invokes undefined behavior however that is not the case.
First let's format the code correctly:
#include<stdio.h>
int main(){
int k = 0;
while(+(+k--)!=0)
k=k++;
printf("%d\n", k);
return 0;
}
So now we can see that the statement k=k++; is inside of the loop.
Now let's trace the program:
When the loop condition is first evaluated, k has the value 0. The expression k-- has the current value of k, which is 0, and k is decremented as a side effect. So after this statement the value of k is -1.
The leading + on this expression has no effect on the value, so +k-- evaluated to 0 and similarly +(+k--) evaluates to 0.
Then the != operator is evaluated. Since 0!=0 is false, the body of the loop is not entered. Had the body been entered, you would invoke undefined behavior because k=k++ both reads and writes k without a sequence point. But the loop is not entered, so no UB.
Finally the value of k is printed which is -1.
Here's a version of this that shows operator precedence:
+(+(k--))
The two unary + operators don't do anything, so this expression is exactly equivalent to k--. The person that wrote this most likely was trying to mess with your mind.
I have a doubt about the for loop of the following code in C:
main()
{
int i=1;
for(;;)
{
printf("%d",i++);
if(i>10)
break;
}
}
I saw this code in a question paper. I thought that the for loop won't work because it has no condition in it. But the answer says that the code has no error. Is it true ? If true, how ?
The regular for loop has three parts:
Initialization
Condition
Increment
Usually they are written like this:
for (initialization; condition; increment) { statements }
But all three parts are optional. In your case, all parts are indeed missing from the for loop, but are present elsewhere:
The initialization is int i=1
The condition is if (i>10) break
The increment is i++
The above code can be equivalently written as:
for (int i=1; i <= 10; i++) {
printf("%d", i);
}
So all the parts necessary for a for loop are present, except they are not inside the actual for construct. The loop would work, it's just not a very readable way to write it.
The for (;;) loop is an infinite loop, though in this case the body of the loop takes actions that ensure that it does not run forever. Each component of the control is optional. A missing condition is equivalent to 1 or true.
The loop would be more clearly written as:
for (int i = 1; i < 11; i++)
printf("%d", i);
We can still debate whether the output is sensible:
12345678910
could be produced more easily with:
puts("12345678910");
and you get a newline at the end. But these are meta-issues. As written, the loop 'works'. It is syntactically correct. It also terminates.
You are not specifying any parameters or conditions in your for loop, therefore, it would be an endless loop. Since there is a break condition based on another external variable, it would not be infinite.
This should be re-written as:
for (int i = 1; i <= 10; i++)
printf("%d",i++);
It's an infinite loop. When there is not a condition in for and we use ;; the statements in the body of for will be executed infinitely. However because there is a break statement inside it's body, if the variable i will be greater than 10, the execution will be stopped.
As it is stated in MSDN:
The statement for(;;) is the customary way to produce an infinite loop which can only be exited with a break, goto, or return statement.
For further documentation, please look here.
Even if a for loop is not having any condition in it, the needed conditions are specified inside the for loop.
The printf statement has i++ which keeps on increasing the value of i and next we have if statement which will check if value of i is less than 10. Once i is greater than 10 it will break the loop.
In the program I have pasted below, I was just wondering why the pointer "p" was initialized within the for loop? I am used to reading the conditions of a for loop as: starting from this value of a variable; until it reaches this value; increment it by this much. So it seems strange to have another variable that does not determine the ending condition and is not being incremented during each iteration in there at all.
I would have just put p=&a[0]; above the for loop and left the rest. Is this just a stylistic thing or are there differences in the way things are processed depending on where you initialize p? Is one way preferred over the other?
#include <stdio.h>
#define PRD(a) printf("%d", (a) )
int a[]={0, 1, 2, 3, 4};
int main()
{
int i;
int* p;
for (p=&a[0], i=0; i<=4; i++) PRD(p[i]);
return 0;
}
This appears to be just a style thing. I would probably have also put the initialisation of p outside the for statement, since cramming everything in there makes the code harder to read. (Because the pattern of that for loop is different from what you might usually expect, an experienced programmer will have to stop, back up, and think about what's in there before it will make sense. I initially thought there were four clauses in the for control statements until I noticed that the first separator was a comma.)
Writing the code like this (instead of initialising p outside the loop) will have no effect on performance.