python to C conversion error

Question

python code

for b in range(4):
 for c in range(4):
 print myfunc(b/0x100000000, c*8)

c code

unsigned int b,c;
for(b=0;b<4;b++)
 for(c=0;c<4; c++)
 printf("%L\n", b/0x100000000);
 printf("%L\n" , myfunc(b/0x100000000, c*8)); 

I am getting an error saying: error: integer constant is too large for "long" type at both printf statement in c code. 'myfunc' function returns a long. This can be solved by defining 'b' a different type. I tried defining 'b' as 'long' and 'unsigned long' but no help. Any pointers?

---

My bad...This is short version of problem

unsigned int b;
b = 1;
printf("%L", b/0x100000000L);

I am getting error and warnings: error: integer constant is too large for "long" type warning: conversion lacks type at end of format warning: too many arguments for format

• Seth Johnson: You're missing some curly braces in your C code.
• GWW: For long integer constants use 0x100000000L
• Random832: What output are you expecting here? Any integer type divided by a larger value is just going to be zero.

Answer 1

Your C code needs braces to create the scope that Python does by indentation, so it should look like this:

unsigned int b,c;
for(b=0;b<4;b++)
{
 for(c=0;c<4; c++)
 {
 printf("%L\n", b/0x100000000);
 printf("%L\n" , myfunc(b/0x100000000, c*8)); 
 }
}

Answer 2

Try long long. Python automatically uses number representation which fits your constants, but C does not. 0x100000000L simply does not fit in 32-bit unsigned int, unsigned long and so on. Also, read your C textbook on long long data type and working with it.

Answer 3

unsigned int b,c;
const unsigned long d = 0x100000000L; /* 33 bits may be too big for int */
for(b=0;b<4;b++) {
 for(c=0;c<4; c++) { /* use braces and indent consistently */
 printf("%ud\n", b/d); /* "ud" to print Unsigned int in Decimal */
 printf("%ld\n", myfunc(b/d, c*8)); /* "l" is another modifier for "d" */
 }
}