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#include <iostream> using namespace std; int main() { int a[]={3,5,2,1}; for(int i=1;i<5;i++) { a[0]=a[i]+1; } cout<<a[0];
its output is 5...how?? can somebody provide dry run for it??
10 Respostas
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as i tried some checks
for i<2 a[0]=6
for i<3 a[0]=3
for i<4 a[0]=2
for i<5 a[0]=5
upto 4 i got how values come but after 4 i couldn't evaluate it
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it is can not be 5. when i = 4 it is out of the massive
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@maria milovanova
run the code and check it
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I checked it. There is an error
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#include <iostream>
using namespace std;
int main() {
int a[]={3,5,2,1};
for(int i=1;i<5;i++)
{
a[0]=a[i]+1;
}
cout<<a[0];
return 0;
}
run this one
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I know that there should be "return 0". As I said, there is an error in the conditions of the loop. Array starts with 0, the size of this massive is 4. And that is why you can not initialize element a[4], it is more then the arrays size. P.S. sorry for mistakes. I don't know english well
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that's the problem in this code. Even though array size is 3 ,how can i goes upto 5??? 😎
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even if conditions will changed on i<4, it cannot be 5, it should be 2
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it is because, when you out of the massive, it adds 1 to num, thet lies beyond the array
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actually the on mobile app this code produces answer as 5 but on pc it produces the garbage values. 😂