C Programming - Arrays - Discussion
Discussion Forum : Arrays - Point Out Correct Statements (Q.No. 3)
3.
Which of the following statements are correct about the program below?
#include<stdio.h>
int main()
{
int size, i;
scanf("%d", &size);
int arr[size];
for(i=1; i<=size; i++)
{
scanf("%d", arr[i]);
printf("%d", arr[i]);
}
return 0;
}
Answer: Option
Explanation:
The statement int arr[size]; produces an error, because we cannot initialize the size of array dynamically. Constant expression is required here.
Example: int arr[10];
One more point is there, that is, usually declaration is not allowed after calling any function in a current block of code. In the given program the declaration int arr[10]; is placed after a function call scanf().
Discussion:
8 comments Page 1 of 1.
Amrutha said:
7 years ago
@All. Please help me.
When I compile my program unexpected error is given but the answer is;
The code is erroneous since the statement declaring an array is invalid.
variable array size is allowed.so, they give segmentation faults.
When I compile my program unexpected error is given but the answer is;
The code is erroneous since the statement declaring an array is invalid.
variable array size is allowed.so, they give segmentation faults.
Noel said:
7 years ago
No, The error is initializing the array at 1 instead of 0. If that is corrected it runs smoothly.
#include<stdio.h>
int main()
{
int size, i;
scanf("%d", &size);
int arr[size];
for(i=0; i<=size; i++)
{
scanf("%d\n", &arr[i]);
printf("%d\t",arr[i]);
}
return 0;
}
#include<stdio.h>
int main()
{
int size, i;
scanf("%d", &size);
int arr[size];
for(i=0; i<=size; i++)
{
scanf("%d\n", &arr[i]);
printf("%d\t",arr[i]);
}
return 0;
}
AKSHAY KALRA said:
8 years ago
Variable sized arrays are allowed.
The only wrong part in this question is for loop and scanf statement. It should be:-
for(i=1; i < size; i++)
{
scanf("%d", &arr[i]);
printf("%d", arr[i]);
}
The only wrong part in this question is for loop and scanf statement. It should be:-
for(i=1; i < size; i++)
{
scanf("%d", &arr[i]);
printf("%d", arr[i]);
}
Shubham_K said:
10 years ago
The code compiles successfully on GCC. But on running, it throws segmentation fault just because of the line 'scanf("%d", arr[i])'.
On replacing it with 'scanf("%d",(arr+1))' eliminates the segmentation fault too, and the code runs successfully.
On replacing it with 'scanf("%d",(arr+1))' eliminates the segmentation fault too, and the code runs successfully.
Sireesh said:
1 decade ago
This program compiles without any errors on gcc compiler. It gives segmentation fault while running because 'scanf("%d", arr[i])' statement may access invalid memory location.
Mitali said:
1 decade ago
None of the option is correct. Code is erroneous just because in scanf '&' operator is not used. Else code works fine.
Meenu said:
1 decade ago
Variable sized arrays are allowed. This gives segmentation fault. So correct answer is:
A The code is erroneous since the subscript for array used in for loop is in the range 1 to size.
A The code is erroneous since the subscript for array used in for loop is in the range 1 to size.
Bhath said:
1 decade ago
#include<stdio.h>
int main()
{
int size, i;
scanf("%d", &size);
int arr[size];
for(i=1; i<=size; i++)
{
scanf("%d", &arr[i]);
printf("%d", arr[i]);
}
return 0;
}
this working in gcc compiler
int main()
{
int size, i;
scanf("%d", &size);
int arr[size];
for(i=1; i<=size; i++)
{
scanf("%d", &arr[i]);
printf("%d", arr[i]);
}
return 0;
}
this working in gcc compiler
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers