Cant access!!
Is it in python?
A segfault basically means you did something bad with pointers. This is probably a segfault:
char *c = NULL;
…
*c; // dereferencing a NULL pointer
Or this:
char *c = “Hello”;
…
c[10] = ‘z’; // out of bounds, or in this case, writing into read-only memory
Or maybe this:
char *c = new char[10];
…
delete [] c;
…
c[2] = ‘z’; // accessing freed memory
Same basic principle in each case - you’re doing something with memory that isn’t yours.
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no, it is c…this is the problem
and this the code
#include <stdio.h>
long long int a[1000006];
int main()
{
long long int n,k,i,co=0,max=0;
scanf("%lld %lld",&n,&k);
for(i=1;i<=n;i++)
{
scanf("%lld",&a[i]);
if(max<a[i])
max=a[i];
}
long long int c[max+1];
for(i=0;i<=max;i++)
c[i]=0;
for(i=1;i<=n;i++)
c[a[i]]++;
for(i=1;i<k;i=i+2)
{
if(c[i]>0)
co++;
}
printf("%lld\n",co);
return 0;
}
thank u…
Why do I get SIGSEGV error for this code http://www.codechef.com/viewsolution/7280734
Please help, I haven’t used pointers or functions and it runs perfectly well on code blocks
#include<stdio.h>
int main()
{
int i,n,f=1,j=1;
scanf("%d",&n);
int a[10];
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
{
f=1;j=1;
while(j<=a[i]){
f=f*j;
j++;
}
printf("%d\n",f);
}
return 0;
}
why am i getting a run time error here?
Can anyone tell me why i am getting run time error(sigsegv) for this
https://www.codechef.com/viewsolution/7491260
Just check if you are accessing any element outside the declared array size,most common error might be in a loop.