Comments by Member 13188016

Member 13188016 16-May-17 3:52am View

no:/
https://ibb.co/gw1tAk
tis is the output of the problem that i want to write
since im beginner so maybe approach this problem in beginner way too :/

Member 13188016 16-May-17 3:50am View

thanks for the answer
https://ibb.co/gw1tAk please refer this link for the output.
to produce this output, is it better use for loop or array(?)
and please explain the way too!!

Member 13188016 14-May-17 8:46am View

thankyou so much!!!!!!!!!!!!!!!!!!!!!!!!! for the detailed explanation!!!!!!!!!!!!!! hope to find someone like you again everytime im confuse!!!lol ><

Member 13188016 14-May-17 8:33am View

ah got it!!!!
it means i=2 because the initialization
and j =2
because the condition is true 4<10000 it will execute the execution statement
then j will be incremented to 3
so i*j<10000 -> 6 <10000 that is why all the value of even number in array become 0
then for i=j=3 it will make the value of multiple 3 become 0 in array
and so on?

Member 13188016 14-May-17 8:14am View

Still dont understand the inner loop part
For (j=i;i*j<10000;j++) {prime(i*j)=0}
Means i=j=2
2^2=<10000(means if the condition true will process the latter or will increment j++ until it reach 10000?);j++
And if j is incremented by 1 it should be 3*3=9 so why the even number all was eliminated? Is the elimination occur in the first iteration?

Member 13188016 14-May-17 8:09am View

I dont understand the inner loop part
For(j=i;j*i<100000;j++)
{prime[i*j]=0}
It start with i=2
J=2;2*2<10000(is this mean if 4<10000 true will process the code or is it mean that it will increment until reach 10000?); j++
How it remove the even number still dont get it

Member 13188016 14-May-17 8:04am View

I dont understand the inner loop here in first iteration it make all even number become zero?
For(j=i;i*j<10000;j++) {prime[i*j]=0}
Index i=2 when enter the loop
I=j=2
2*2=4<10000
J++
This mean start j=2 then multiply it j=2*2 if it under 10000 then increment j with 1, so after it is 3*3=9(?) i dont understand this part!!

Member 13188016 13-May-17 14:38pm View

Can i see return value in debugger?
How to debug at machine language level?
Because some debugger only allow to watch value of variable

Member 13188016 13-May-17 9:31am View

Ah sorry the correct one is int_bits (void) {
Return count_bits(~0U)
}
But still dont understand how it produce 1010 for 10
And what to compare to make it produce 1010

Member 13188016 13-May-17 9:31am View

Already change my questions! can you explain and answer my question?

Member 13188016 9-May-17 22:47pm View

Thankyou for the recommendation!!! Will read it for better understanding!!

Member 13188016 9-May-17 22:46pm View

Thankyou!!!!!!! So the pointer data type determine its value?
For example char *my_pointer so the value always only one word like the first word of an array(?)
If int *my_pointer then the value is number?
For an array *my_pointer obtain value where pointer is aimed which is the first term of an array so that's why the data type is always char for an array?

Member 13188016 9-May-17 1:26am View

Thankyou for answering my questions!!!
But still confuse in some part, So the value of where pointer aimed in array always the first item of an array, But actually inside the poiner is it contains all the words address or only the first item address?
It means that *my_pointer only contains the address of first word?
While my_pointer refer to the address of whole words?
Is there any relation with data type char too?

char *my_pointer;
my_pointer = string;
Since my_pointer is a pointer to char, the value of my_pointer is an address which points to char.
Data type char only refer to one word like 'h' for example or all words?

Member 13188016 9-May-17 1:25am View

Deleted

So the value of where pointer aimed in array always the first item of an array, But actually is it contains all the words or only the first item address?
What is the different if i use my_pointer in printf function? And why it crash if use *my_pointer with %s ?

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