How to interpret *( (char*)&a )

Question

I see a way to know the endianness of the platform is this program but I don't understand it ``` #include <stdio.h> int main(void) { int a = 1; if( *( (char*)&a ) == 1) printf("Little Endian
"); else printf("Big Endian
"); system("PAUSE"); return 0; } ``` What does the test do?

Accepted Answer

An `int` is almost always larger than a byte and often tracks the word size of the architecture. For example, a 32-bit architecture will likely have 32-bit ints. So given typical 32 bit ints, the layout of the 4 bytes might be:

```
   00000000 00000000 00000000 00000001
```

or with the least significant byte first:

```
   00000001 00000000 00000000 00000000
```

A char* is one byte, so if we cast this address to a char* we'll get the first byte above, either

```
   00000000
```

or

```
   00000001
```

So by examining the first byte, we can determine the endianness of the architecture.

How to interpret ( (char)&a )

Solution

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