I have an int variable that contains an hex number like this 084c2365. And I want to convert the first byte(65) into a decimal number (101).
But I get an error at compilation the says that the 1st argument from sscanf is not an "const char *".Code:char c[100]; int num; //Num has the value 084c2365 int result; c[0]=(char)num; //c[0] gets the value from the firts byte (65) sscanf(c[0], "%x", &result);
How can i fix this??
Thanks.
c[0] is a char.
initialize your variables.
decimal or hexdecimal - is just representetion during output - internally - it is the same number stored as binary, no need to convertCode:int num = 0x084c2365; int result = num & 0xFF; printf("%d", result);
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
Thanks, but does it matter if I have my num variable as an uint32_t and I just have the 84c2365 number without the 0x notation?
So will
make the conversion from hex to decimal??Code:int result = num & 0xFF;
Sorry I am a junior and I need to learn a lot.
Thanks
If you have 10 apples or 0x0A apples or 012 apples, how many apples do you have?
The whole idea of 'base' is just a presentation issue, so whether you say 10, 012 or 0xA doesn't matter, the underlying number is still the same.
Hex is just convenient since 2 hex digits fit exactly in one 8-bit byte, so from that point of view, it's a bit easier to imagine what is going on when you say
int result = num & 0xFF;
rather than
int result = num & 255;
But the result is exactly the same as far as the machine is concerned.
Likewise, octal was great on certain historic machines which had say 6-bits or 9-bits per byte.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
