Question

As far as I know, in C programming language, an array is stored on the memory element by element. (i.e., element 0, element 1, element 2, … , element n). I’m trying to see that with the following code:

unsigned char a[] = { '\1' , '\2', '\3' ,'\4' };  
printf("%d\n", (int*) a);

Since unsigned char is 1 byte and an integer is 4 bytes; I thought it has to print the value:

00000001 00000010 00000011 00000100 = 2^2 + 2^8 + 2^9 + 2^17 + 2^24 = 16909060

However, when I run this program, it generates different results for every trials.

What am I missing here?

Answer 1

You probably want to use *(int *)a, otherwise you’re just printing an address.

However, this will invoke implementation-defined behaviour:

• You will get a different result depending on the endianness of your platform.
• Depending on the platform, the char array may not be properly aligned to be read as an int.
• The compiler may perform funky optimizations based on assumptions that you will never read the char array through an int * – you are breaking what are known as the strict aliasing rules.