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isn't it error, its trying to access a[99][0] here initially?

I think it will give you some random answer. It is checking for the element at position a[99][j] in a loop, now it depends that which element is at that memory location. If ever in the whole loop that element comes out to be smaller than k then k's value will be updated else it will print 99.

i'm getting zero...

here the output is '0' or any garbage value,why because a[k][j] means a[99][j] means any garbage value which may be greater than 'k' or less than 'k' so "a[k][j]>k" may be true or false.(we may get output '0' in some cases because some values of array are not intialised ,by default they are '0').

segmentation fault

Undefined behavior. I don't think segmentation fault would occur because the 2d array is allocated dynamically on the stack, not on the heap through malloc(). Could be wrong about this though.

It depends on the garbage values stored at a[99][0], a[99][1], a[99][2], and a[99][3]. If none of these values are less than 99, then k never gets updated. Otherwise it gets set to a[i][j]. Since 4<=a[i][j]<=12 and there are only 3 rows in the matrix, the new value of k will still produce garbage values.

99

99

Answer is undefined. The values up to the end of the array a[3][4] , i.e upto 12 values will be 0 if it is uniniitalized .However any value beyond that can give a runtime error as it will have a garbage value . The statement a[k][j] will make the array access unpredictable and hence cause a runtime error . I tested with vc++ a[99][j] had -8567433 so the program gave a runtime error .

i am getting nine...... and if i replace 9 in the array with ne othr numbr then m getting 0.....

99 obviously!! Since it will execute only last line "printf" since all others are a part of for in which condition is not satisfied.

main()
{
int a[3][4] ={1,2,3,4,5,6,7,8,9,10,11,12} ;
int i, j , k=99 ;
for(i=0;i<3;i++)
for(j=0;j<4;j++)
if(a[i][j] < k) k = a[i][j];
printf("%d", k);