c++ with unix?


 
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# 1  
Old 09-17-2008
c++ with unix?

im quite a noob in both unix and c++ and i encountered this line

lc = 41
i think i = 1 first since this line is included in a for loop
sprintf(like,"%s%i","0",lc);

I cant understand this line

when it is first executed the output is
041
# 2  
Old 09-18-2008
Code:
lc = 41
i think i = 1 first since this line is included in a for loop
sprintf(like,"%s%i","0",lc);

the first line, the variable named 'lc' is assigned the number 41
The sprint() function prints the combination of a string (%s) and an integer (%i) which are "0" (zero as a string) and the integer which was assigned above to be 41
Therefore the output is "0"+"41" or 041
# 3  
Old 09-18-2008
I see
so why do the 0 got printed first?

and this one too

if(bd%3==0)

i cant understand this line too
can you help me with this too?
# 4  
Old 09-18-2008
oh nevermind my question
just help me with this code

if(bd%3==0)

why is there %3?
 
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