What is ouput will be printed in array , if following code performed

Question

Main () { 
int a [3] [4] = $\begin{pmatrix}
1&2&3&4 \\
5&6&7&8 \\
9&10&11&12 \\ 
\end{pmatrix}$ 
printf ("\n% u% u% u", a[0]+1, * (a[0] + 1), 
*(*(a + 0)+1)); 
}

What is the output of the above program? Assume array begin at address 10.                                                    

Comments

when i excute 3rd statement of printf then what is the output???

anyone explain in detail...

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a+0 means we are skipping 0 rows. Means we are not skipping any thing. (a+0)+1 means we are skipping only 1element in row 1, so we move to 2. It is same as statement 2 but different representation.

Answer 1

12, 2 , 2

first will print address of 1st row 2st column element  which is 10 +2= 12

second will print value at that address i.e. 2

Third is same as second i.e. different expression = 2

Comments

12 or 14 sorry depends on size

//

Answer 2

a[0]+1

Given code will skip 1 element of 2d array so it will print address of second which is 14.(Given a=10).

a[0]+1

Will print value at the address (a[0]+1) which 2.

*(a[0]+1)=*(*(a+0)+1)

Same as above Will print value at the address (a[0]+1) which 2.

SO output:14,2,2

Comments

Yes a+1 mean 1 row will skipped

*a+1 after selecting row 1 element will be skip

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Thx...

So I think it should be 12 not 14 ...as it moving to 2 and type unsigned int, which will increment the address by 2....if we consider row major order.

And there is nothing mention about ordering, so how to figure out which order we should consider.

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12 or 14 depends on size of int.

by default row major otherwise it will be mentioned.