As the book said, you can certainly copy a structure like this:
WordCopy[i]=Word[i];
Maybe you copied the array of structures instead of the structure itself like this:
WordCopy=Word;
Of...
Type: Posts; User: orientuser
As the book said, you can certainly copy a structure like this:
WordCopy[i]=Word[i];
Maybe you copied the array of structures instead of the structure itself like this:
WordCopy=Word;
Of...
no difference
int array[][] = {0, 1, 2, 3, 4, 5, 6, 7, 8};
It's grammatically incorrect.
cas is right.
"I want to be able to pass arrays of float values as well as arrays of double values to the function."
-- If you really want to do so, you must use function overriding in C++.
...
Since the arrays are local, i.e. declared in a function, you must pass them as arguments to your new function.
I think the following code can work as you expect.
for (i=0; i < numoflines;...
output is 0
May I ask you some questions?
1
Are the arrays, such as year, month, local or global?
2
since one of the 'if' conditions is if (year[j] == year[j+1])
why did you exchange the values of...
If mec is an auto local variable(no static precedes volatile unsigned int), they are identical.
sorry, i made a mistake. In the 1st code snippet, args is a pointer to struct aggregator_t, not a structure variable.
memcpy (x, (const void *)&y, 2);
& is redundant.
The binary code of one of the 2 bytes in y is 10010110. If this byte is regarded as a signed char, its value is just -106.
BTW: A plain char might be regarded as signed char or unsigned char,...
They are different.
In the 1st code, a structure variable named args is declared and then args is an object. In the 2nd code, however, args is a new alias for the type "struct aggregator_t *" .