Question

Explain the following code at each line. show a struct and with arrows indicate the operations that are being performed. give as much detailed explanation as possible for the following code in C .

//print 1st matrix
printf("\n1st matrix:\n");
i=0;
while(i<3){
j=0;
while(j<3){
printf("%4d", *(*(a+i)+j));
j++;
}
i++;
printf("\n");
}

Answer 1

//print 1st matrix
printf("\n1st matrix:\n");
i=0;
while(i<3){
j=0;
while(j<3){
printf("%4d", *(*(a+i)+j));
j++;
}
i++;
printf("\n");
}

The main part in the above code is identifier 'a'. It on a whole represents a 2-Dimensional array but the value in the variable 'a' has address of the 0th indexed 1-D array(Also called the base address). So, on adding 1 to a (a+1), we will be pointing to 1st indexed 1-D array.

By using unary operator astreik, *, the current pointing will be done to 0th element address of nth 1-D array. For example, *(a+1) points to 0th element in the 1st indexed 1-D array of the matrix. So, again using the same operator on the above statement will give the value of 0th element in the 1st 1-D array of the matrix. For example, *(*(a+1)+2) will give the value of 2nd element of the 1st indexed 1-D array in the matrix.

Let the given matrix is having values as below:

5 7 3

9 3 2

4 8 6

Now coming to the program, i will iterate from 0 to 2 and similarly j also but inside i loop iteration. So, the expression *(*(a+i)+j)) inside the printf statement will be accessed 9 times(3*3) as follows,

i = 0

*(*(a+0)+0)), gives value of 0th index of 0th 1-D array(1st row of matrix) , i.e., prints 5

*(*(a+0)+1)), gives value of 1st index of 0th 1-D array(1st row of matrix) , i.e., prints 7

*(*(a+0)+2)), gives value of 2nd index of 0th 1-D array(1st row of matrix) , i.e., prints 3

i = 1

*(*(a+1)+0)), gives value of 0th index of 1st 1-D array(2nd row of matrix) , i.e., prints 9

*(*(a+1)+1)), gives value of 1st index of 1st 1-D array(2nd row of matrix) , i.e., prints 3

*(*(a+1)+2)), gives value of 2nd index of 1st 1-D array(2nd row of matrix) , i.e., prints 2

i = 2

*(*(a+2)+0)), gives value of 0th index of 2nd 1-D array(3rd row of matrix) , i.e., prints 4

*(*(a+2)+1)), gives value of 1st index of 2nd 1-D array(3rd row of matrix) , i.e., prints 8

*(*(a+2)+2)), gives value of 2nd index of 2nd 1-D array(3rd row of matrix) , i.e., prints 6

Note: All the printed value in the output will length of 4 because we have %4d in printf(). For example, to print value '6', It prints as '' 6". It has length 4.

Hoping that the explanation helps.