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I am expecting a to be 5, making the comparison expression evaluate to true and thus print:
A = 5, B = 3, C = 6
main()
{
int i,a,b=3,c=6,s;
for(i=1;i<=1000;i++){
if(a*a+b*b==c*c){
printf("A = %d B = %d C = %d\n",a,b,c);
}
a = i;
}
}
Where is a initialized? I am seeing i being set, as well as b and c, but I'm not seeing any assignment to a.
The solution is to assign the current value of i to a: a=i prior to the if statement. You are assigning the value after the fact. Alternatively, evaluate i, not a.
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151A codeforces link
Here is my code written in C language. It has some problem
#include<stdio.h>
#include<conio.h>
int main ()
{
int n,k,l,c,d,p,nl,np,mm,per,tl,to,sum;
scanf("%d%d%d%d%d%d%d%d",&n,&k,&l,&c,&d,&p,&nl,&np);
mm = k*l;
per = mm/nl;
tl = c*d;
to = p/np;
sum=(per,tl,to)/n;
printf("%d",sum);
return 0;
}
You have to find the minimum value among per, tl, and to and divide that with n instead of just calculating to/n, which you are currently calculating with sum=(per,tl,to)/n;. (per and tl are ignored according to the definition of the comma operator)
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In the main() function,
i,a and b are decalred static. Ok, fair enough.
The function is called.
All i,a and b have some value. Ok, fair enough.
The function is called again.
Now as a static variable i , a and b must retain their value.
But
How is i again intialised to 0? (Shouldnt it contain its previous value?)
Snap shot of the problem.
P.s Answer is d btw.
i is not initialized again:
void printtab()
{
static int i, a = -3, b = -6;
i = 0;
...
It is assigned a new value when the function is entered.
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It is said that,in c,b++; is equal to b=b+1; if this is the fact test++ in my code why generate a compile time error.
test+1 is working well but test++ is not working.but why?
#include<stdio.h>
int main(void)
{
char test[80]="This is a test";
int a=13;
for(;a>=0;a--)
{
printf("%c",*(test++);
}
}
The ++ and -- operators are not defined for arrays.
v++; would be the same as v = v + 1;. Assumed v was typed an array this would imply assigning to an array, which is not defined.
char test[80] = "This is a test";
char *p = test;
for(int a = 0; a < 14; a++)
{
printf("%c", *(p++));
}
Well, for one thing, b++ is not the same as b=b+1.
But even if it were -- I think you'll find you get a similar error if you try test = test + 1.
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Here is the source code：
#include <stdio.h>
enum coordinate_type{ RECTANGULAR = 1,POLAR };
int main(void)
{
int RECTANGULAR;
printf("%d %d\n",RECTANGULAR,POLAR);
return 0;
}
Why is the result the following:
3 2
You are redefining what RECTANGULAR is in the main function. It gets initialized with a "random" value, in this case it is 3, but it could be anything else.
POLAR keps its value of 2 because of how the enum is defined.
Try redefining the RECTANGULAR variable in main to see different outputs.
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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".