## Jan 22, 2013

### C Aptitude: Endianness, Pointer Arithmetic

C Aptitude 31

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Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

Predict the output or error(s) for the following:

C aptitude 31.1
```
main() {
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
```

Explanation: The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the INTEL machines are ‘small-endian’ machines. Small-endian means that the lower order bytes are stored in the higher memory addresses and the higher order bytes are stored in lower addresses. The integer value 258 is stored in memory as: 00000001 00000010.

C aptitude 31.2
```main() {
int i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}
```

Explanation:The integer value 300  in binary notation is: 00000001 00101100. It is  stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it  is  00000010 00101100 => 556.

C aptitude 31.3
```main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
least = ((*ptr)<(least))?(*ptr):(least);
printf("%d", least);
}```
Explanation: After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is ‘\0’. So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.
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