Why doesn't the compiler show an error for *marks being used as marks[I]?

Question

1.00/5 (1 vote)

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#include <stdio.h>
int main()
{
int *marks;
printf("Enter four marks:");
for(int i = 0;i<4;i++)
scanf("%d",&marks[i]);
printf("Marks you entered are:");
for(int i=0;i<4;i++)
printf("%d ",marks[i]);
return 0;
}

What I have tried:

The program did not give any output neither did it show any error

Posted 2-Oct-18 8:24am

Kritika Choudhary

Updated 2-Oct-18 9:27am

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W∴ Balboos, *GHB* · Answer 1 · 2018-10-02T08:43:00

Solution 1

You declared
int *marks;

You didn't allocate any memory for it. You're somewhat fortunate your application didn't just crash.

Something like
marks = (int *)malloc(4*sizeof(int));

Posted 2-Oct-18 8:43am

W∴ Balboos, GHB

Updated 2-Oct-18 8:44am

v2

Comments

CPallini 2-Oct-18 17:19pm

5.

OriginalGriff · Answer 2 · 2018-10-02T09:28:00

Quote:
Why doesn't the compiler show an error for *marks being used as marks[I]?

Because the language definition says that the name of an array is a pointer to the first element. So marks is both a pointer and an array t the same time; the usage is available either way you declare it.

C++

int arr[10];
int i1 = arr[0];
int i2 = *arr;

Is all valid, as is:

C++

int *arr = (int*) malloc(40);
int i1 = arr[0];
int i2 = *arr;