So,I understand everything in code except this if statement - if(i-2) :
for (i = -1, j = 0; i < 4; i++, j++)
if (i - 2)
printf("%d\t", j * (i + 2));
Can anyone explain meaning of this if statement,does it have any influence on printf ? Thanks.
The statement under an if is executed if the controlling expression is true. In C, an expression that evaluates to 0 is considered false, while any other value is considered true.
So if (i-2) means the printf will get executed if i-2 is non-zero, i.e. when i is not equal to 2.
In C, an integer is treated like a boolean in an if statement:
0 is false
not-zero is true
So if (i-2) is equivalent to if (i != 2).
In c any value can be used in an if statement not just boolean expressions. This devolves into expression != 0. Often this is against style guides as it can be confusing and error prone.
The two following are equivalent
if (val) ...
if (val != 0) ...
So in you're case
if (i - 2)
will be true for every value of i except 2 when i-2 is 0
Related
So I saw this in my text book and I cannot understand how it works.
x=0;
int i = 0;
int n;
while(!x && i < n){
if(array[i]==target)
x=1;
else
++i;
}
what I don't understand is how "!x" works in loop's condition.
what I understand is this loop keeps running until it runs n times or when array[i]==target, which will change the value of x = 0 to x = 1 and stop the loop.
what I tried:
I tried replacing !x to x==0 and it did the same job.
The short answer here is that it does not matter what !x is because it is surrounded with undefined behavior completely nullifying its effect.
In the statement
While(condition) {...}
condition must resolve to either true or false, and must therefore be logical expression.
Given:
int x=0;//This answer assumes 'int' here as not specified in OP
int i = 0;
int n;
while(!x && i <n)
Because x is initialized as 0, equivalent to false the expression !x resolves to true, satisfying part of the condition.
Because n is not initialized however, the condition (!x && i <n) invokes undefined behavior, making the results of the overall condition unknown at the time of first entry, i.e. it can be either true or false. Further more, n does is never modified within {...}, so if the initial path happens to enter the brackets, the condition will change only due to changes in i. But again, because the value of n is not known, only undefined behavior will occur.
I don't know what language this is but !someVar is typically shorthand syntax for "variable is false", or another way of writing someVar == false. And in binary 0 is false and 1 is true. In the C language, 0 is false and anything not 0 is true. Therefore, in your loop, while !x is shorthand syntax for while x == false or while x == 0. Again, generally speaking since we don't know the language.
While(!x) means if x is a condition then we go into the while loop, only if the condition does not satisfy.Here in your code you have assigned x as zero,so !x means 1 because '!' Basically means opposite if x =0 then !x =1, if x=1 then !x=0 (all numbers greater than 1 are also considered as 1 only) so while(1 && i<n) is what happening here in the next step.If i<n is true then you will enter into the loop. If i<n is false the you won't enter the loop.
while(!x)
{
intructions;
}
means that your instructions will only keep running over and over again if x is false or equals 0 and will stop executing instructions once x becomes true or different from 0
I've seen the operator ! being used in multiple places differently and I still don't get how it actually works. My basic understanding is it reverses the value from true to false and vice versa. If it reversed to true the statement triggers. Let's take an example.
int main(void)
{
int a = 5;
if (!(a == 6))
{
printf("unlike\n");
}
if (!(a == 5))
{
printf("like\n");
}
}
In the code above since a is 5 it ends up printing "unlike" because the false statement that a is 6 got reversed. Now let's take another example.
int main(void)
{
string i = "abc";
string j = "cab";
string k = "abc";
if (!strcmp(i, j))
{
printf("unlike\n");
}
if (!strcmp(i, k))
{
printf("like\n");
}
}
The string type has been taken from the cs50.h header and strcmp from string.h. strcmp returns value 0 if the two strings are alike and if unlike depending on the alphabetical order returns a positive or negative value. Now if we follow the logic in the previous example, since i and j are unlike, and false it should be reversed to true and unlike should be the output. But I tried running the code and the result was like.
I am confused. Can anyone please explain this to me clearly? Feel free to use other examples too. I could always get away with not using ! but I just want to learn what it is and how to properly use it.
A boolean in C is an integer with zero for false and non-zero for true.
strcmp returns 0 when the compared strings are identical and a non-zero value depending on the difference otherwise. Therefore, strcmp(i,k) is seen as "false". The ! then changes this to "true", which leads to your current output.
In the first case a = 5. then if (!(a == 6)); here a = 6 is not true (false), so it's something like this. if (!(false)) it means if (true). That's why it prints "unlike".
strcmp(i, j) returns 0 if the strings i and j match; otherwise, it will return a non-zero value. In your case,
(!strcmp(i, j))
Here i and j are not equal so strcmp will return a non-zero value because i != j. So !(1) means not(1) means 0, so the if condition is false because of zero. Therefore it'll not execute the printf("unlike\n") line.
(!strcmp(i, k))
Here i and k are same so strcmp will return 0. !(0) means not(0) = 1 so the if condition is true. It will execute the printf("like\n") line.
Conditional operator in C is used like this:
condition ? value_if_true : value_if_false
What does 0 mean when it's used in the value_if_false?
I've seen some people using it like this, for example.
a == b ? i++ : 0
It seems like it does nothing. Does this work like return 0 in other functions?
In C language, ternary is shorter version of if statement and it requires both statements, if_true and if_false. It would be like this (in fact it can have multiple statements for one case, separated with comma):
Short:
condition ? if_true : if_false;
Long:
if (condition) {
if_true;
} else {
if_false;
}
You can also assign the value if you put something infront of condition.
Short:
result = condition ? if_true : if_false;
Long:
if (condition) {
result = if_true;
} else {
result = if_false;
}
Now here is the trick. In C language, writing 0; is a valid statement, so your ternary becomes in longer version same as code below:
if (a == b) {
i++;
} else {
0; /* This is valid C statement */
}
Or if you have assignment too, it would be:
if (a == b) {
result = i++;
} else {
result = 0;
}
You can also do this:
int a;
/* Code here ... */
condition ? a = 5: 0;
That is effectively the same as:
if (condition) {
a = 5;
} else {
/* DO not touch a */
}
The ?: operator is a ternary operator, but it is not called "ternary" as some answers and/or comments here suggest. It just is the arity of the operator, just as + is a binary operator or as & is unary. If it has a name at all, it is called "Conditional Expression"-operator
It is not quite equivalent to if/else, because it is a conditional value (with the consequence, that both expressions must have the same type) in the first place, not a conditional execution. Of course, both types can be cast to make them equal.
In the case of what the OP does, a better option (if if shall not be used) is in my opinion:
a == b && i++;
which resembles a bit more logical what happens. But of course it is a matter of style.
The reason why someone might want to write a == b ? i++ : 0; is that s/he probably wants to have an (Caution! You are now entering an opinion-based area) easier and faster alternative to if (a == b) i++; - although this is of course opinion-based and I personally not share the same opinion.
One thing I can think of as a "blocker" at the if statement is the requirement to write the parentheses () which can be omitted by using the conditional operator instead.
"But why the 0?"
The C syntax requires a third operand for the conditional operator. Else if you would want to compile for example:
a == b ? i++;
you will get an error from the compiler:
"error: expected ':' before ';' token"
Or respectively, doing so:
a == b ? i++ : ;
would raise:
"error: expected expression before ';' token"
So they use 0 as kind of "syntax satisfier" to be able to use the conditional operator as replacement for the if statement. You could use any other numeral value here as well, but 0 is the most readable value, which signifies that it has no use otherwise.
To showcase the use at an example:
#include <stdio.h>
int main (void)
{
int a, b, c = 4;
a = 2;
b = 2;
a == b ? c++ : 0;
printf("%d",c);
return 0;
}
The output for c will be 5, because a == b.
Note that a == b ? i++ : 0 is different when used f.e. inside of an assignment like f.e.:
int c = a == b ? i++ : 0;
Here c is either getting assigned by i or 0, dependent upon a == b is true or not. If a == b is true, c is assigned by i. If a == b is wrong, c is assigned by 0.
Side Notes:
To view it from a technical point, ?= is called the "conditional operator". The conditional operator is one of the group of ternary operators.
If you want to learn more about the conditional operator ?=, look at ISO:IEC 9899:2018 (C18), ยง6.5.15 - "Conditional operator" for more information.
There's nothing special about 0 one could write
a == b ? i++ : 1
And it would behave the same way.
Only difference is when you assign the expression to say another variable:
int c = a == b ? i++ : 1;
// a == b -> c will be i++
// a != b -> c will be 1
However it's much cleaner to write
if (a == b) i++;
It helps to think of the ternary operator as a shorthand way or writing an if-else statement.
If(a == b){
i++;
}else{
//if assigned, return 0. Else do nothing.
}
Can someone explain what the exclamation point in the if statement does (i.e. !strmcp)?
string names[] = {"EMMA", "RODRIGO", "BRIAN", "DAVID"};
// Search for EMMA
for (int i = 0; i < 4; i++)
{
if (!strcmp(names[i], "EMMA"))
{
printf("Found\n");
return 0;
}
}
printf("Not found\n");
return 1;
For an if statement, if the expression evaluates to 0, then the block of code following the if statement is not executed. Any other value (positive or negative), will result in executing the code block. The function strcmp uses 0 to say that strings are equal because less than 0 is used to differ from greater than 0.
So in this code, we want printf("Found\n"); to be executed when the strings are equal. Since strcmp results in 0, we need to negate the value so that it becomes 1 which will result in executing that code block.
strcmp() returns 0 if the strings are identical, so you need to negate it, if you use it in an if clause to assert a true statement.
If your clause is if(0), the code inside the condition will not be executed.
For completion, it returns negative if the first different character found is lower in the first string, for instance:
first parameter string: "abca"
second parameter string :abcd"
This will return negative. If it's the other way arround it will return positive.
Also, string is not usually used in C (I refer you to Jonathan Leffler's commment), you can use char*:
char *names[] = {"EMMA", "RODRIGO", "BRIAN", "DAVID"};
Unary operator ! is called the logical NOT operator (cf., for example, this definition at cppreference.com). ! expression returns 1 if expression evaluates to 0, and it returns 0 if expression evaluates to anything else but 0.
So the condition in if (!0) gives 1; this means, the condition is met and the if-block is entered. It has the same meaning as if(0==0)
Consequently, the meaning of
if(!strcmp(names[i], "EMMA"))
in your code is exactly the same as
if(0==strcmp(names[i], "EMMA"))
And you already know when strcmp returns 0...
The exclamation point is the C's boolean negation character.
It means give me the boolean opposite of the value. A boolean is either true or false, which are 1 or 0 in the C Language.
In C, if statements execute their conditional statements if the argument is true.
if (a) means if a is true (i.e. non-zero)
if (!a) means if a is false (i.e. 0)
Therefore:
if (a) is the same as if (a != 0)
if (!a) is the same as if (a == 0)
Sometimes you'll see code that uses two exclamation points in a row "!!"
For example:
int a = !!b;
That ensures a will be ONLY 0 or 1, regardless of what the value of b is.
If b is ANY non-zero value, the ! operator will treat it as though it is true true, which it treats as being the same as 1
So:
!0 == 1
!1 == 0
!52 == 0
!25692 == 0
The second ! does the boolean inversion again, so:
!!0 == 0
!!1 == 1
!!52 == 1
!!25692 == 1
In C any non zero value is considered ad the logical truth, zero i considered as logical false. ! is a logical negation. So !0 (not false) will be the truth and if(!strcmp(str1,str2)) {statements} statements will be executed when str1 will be same as str2
Code
#include <stdio.h>
int main() {
int i;
for (i=1; i<=10; i++) {
(i % 2) ? printf("%d is odd\n", i) : printf("%d is even\n", i);
}
}
Result
1 is odd
2 is even
3 is odd
4 is even
5 is odd
6 is even
7 is odd
8 is even
9 is odd
10 is even
In the above C program, why it still works fine even though the conditional expression only states i%2 and not i%2!=0 ?
In C, integers can be used in a Boolean context, and zero represents false while non-zero represents true.
That's why your code works. The expression num % 2 will be 0 (the single false value) for an even number and 1 (one of the many possible true values) for an odd number.
The following expressions would all work for detecting an odd number:
num % 2
(num % 2) != 0
((num % 2) != 0) != 0
... and so on, ad untilyougetboredum (like 'ad infinitum' but with limits).
Having said that, I don't really consider it a good idea to do it this way, code should express intent as much as possible and the intent here should be to choose the path of execution based on a comparison. That means, if you're looking for an odd number, you should use something like (num % 2) == 1.
You also don't need a separate printf call in each of those code paths:
printf("%d is %s\n", num, ((num % 2) == 1) ? "odd" : "even");
You'll notice I've also used num instead of i. This is simply a style thing of mine, related to the afore-mentioned intent. If the variable is only used as an index, I'm happy to use the i-type variables(a) but, the second it gains a semantic property (like a number being checked for oddity), I tend to use more descriptive names.
I have no issue with people using simple variable names, I just prefer more descriptive ones in my own code.
(a) Actually, I'd probably use idx in that case but that's being too CDO(b), even for me :-)
(b) OCD but in the right order :-)
C doesn't have a dedicated boolean type. It uses int value as boolean. That is 0 is considered false and any non zero value is treated as true.
Try printing some conditions
printf("%d",5==5);
printf("%d",1>3);
This will output
1 and 0.
C always uses 1 to denote true. But any other non-zero value would work as well when using in conditions.
if(6+1)
printf("TRUE");
Will print TRUE.
This is also the reason we can use this form of while loop:
int i= 10;
while(i--){
printf("%d",i);
}
Will print 9876543210. Notice it stops when i becomes 0, which is false.
Now back to the question, i%2 would always result in either 0 or 1. In case of 1(true) the first statement is run while in case of 0 (false) the second statement is run.