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What does this line of code mean?
if((c-a)!=0 && (d-b)!=0) goto trap;
What does the !=mean?
I inherited this, and I'm not sure what it means.
It means: if c is not equal to a, then if d and b are also not equal, goto (jump to) the line with the label "trap".
The !=, just means not equal. It is the complement of the == operator.
Not equal.
1 != 2
1 == 1
2 != 1
3 != 5
!= means not equal to.
ex: c-a is not equal to 0.
a != b means that a is not equal to b. It's the same as !(a == b).
Here's an english translation of the code...
if c minus a is not equal to 0 AND d minus b is not equal to 0 then goto trap
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Write a single statement to accomplish each of the following. Assume that variables c (which stores a character), x, y and z are of type int, variables d, e and f are of type double, variable ptr is of type char * and arrays s1[100] and s2[100] are of type char.
Determine whether the value of c is a letter. Use the conditional operator to print " is a " or " is not a " when the result is displayed.
printf("%s",((c >= 141 && c<= 172) || (c>= 101 && c<= 132)) ? "is a " : "is not a ");
The C programming has a feature of being case-sensitive. C and c have different ASCII values as characters AND they're different too as identifiers.
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Closed 5 years ago.
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Why output is z=0 in the program written below ?
why isn't it giving any error?
Please Explain
#include<stdio.h>
int main()
{
int x=3,y,z;
y=x=10;
z=x<10;
printf("x=%d, y=%d, z=%d",x,y,z);
return 0;
}
Output :
x=10, y=10, z=0
According to N1570 draft (c11):
6.5.9 Equality operators
The == (equal to) and != (not equal to) operators are analogous to the relational operators except for their lower precedence. Each of the operators yields 1 if the specified relation is true and 0 if it is false. The result has type int.
So in your case, condition is false, because 10 is not less that 10, so 0 is stored to the z.
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What's really happening behind?
int a=10,b=5,c=3;
b!=!a;
c=!!a;
Why the values of b and c are 5 and 1 respectively?
b is 5 because you've assigned it that value and never changed it, since
b!=!a;
...is just a condition that you don't do anything with, not any form of assignment.
c is 1 because a is 10 and !10 is 0, and !0 is 1, thus !!a is 1 (a is 10).
You already have got the answer to your question, but just to have a deeper look into why it happens that way, let me add my two cents here.
An expression like
b!=!a;
is same as
b != !a;
because of the maximal munch rule used in the translation phase. Basically, this says that the longest possible token should be selected from the input while creating a construct (obviously, valid/ meaningful).
Following this principle the ! and = are considered together to form the perfectly valid operator != and the expression is parsed like
b != !a;
over
b ! = !a; //or anything else.
That is why, there is no assignment, as you might have thought.
That said, ! is an unary operator, which does not change the value of the operand. So, taken together, your code is essentially same as
int a=10,b=5,c=3;
c=!!a; //double negation
so, a and b are unchanged, and c is 1 (because !!10 == !0 == 1 )
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I have this code troubling me for a while.The expression at Line 7 is troubling me. Is it giving 0 or -1. If its 0 then answer is 2 else answer is 4.
/* How to find value of c in Line 7 expression */
#include<stdio.h>
int main()
{
int a,b,c=1;
a=b=c; // a,b,c have equal value
c=b+=a=-c; // what will be the output of this expression?
c=-c;
c=(++c)*2;
printf(“%d”,c);
return 0;
}
In C the pre-increment (decrement) and the post-increment (decrement) operators requires an L-value expression as operand. Providing an R-value or a const qualified variable results in compilation error.
An lvalue is a value that can be assigned to.
/* what is value of c */
That's easy to answer: Your question does not make any sense.
Since your code won't compile, c won't have a "value".
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I am wondering how to make it so that if lets say the variable A isnt equal to 1 and 2 do
printf("Unknown letter type again\n");
thanks in advance if you know what to do and also is it possible if you type a letter to make you type a known letter
if (A !=1 || A!=2){
printf(foobar);
}
After you changed your question, the answer would be...
if (A !=1 && A!=2) {
printf(foobar);
}
if (!((A == 1) || (A == 2)))
{
printf("Unknown letter type again\n");
}
if A is not 1 or 2: print message
but if you are working with numbers in text/ASCII, you need to compare A == '1' instead of A == 1.