FamZheng FamZheng 6 6 silver badges 16 16 bronze badges. Why volatile instead of const? I think it is an easy way to avoid any undesired compiler optimization, and actually look at the data in the ram cell to determine the answer.
Or am I missing something here? Hmm not sure This isn't my forte. So I'm not sure. Nate: Yes. Ah yea that would be it. The Overflow Blog. Podcast Making Agile work for data science. Stack Gives Back Featured on Meta. New post summary designs on greatest hits now, everywhere else eventually.
Linked See more linked questions. In which some machines or little endian and some are big-endian. We can easily understand how data is stored in little endian and big endian machines. Let us assume, we have a 32bit processor and need to store a value in memory. Now let us see how the storing order of bytes will depend on the endianness of the system little endian vs big endian. So here we will go first for the Big endian machine and then little endian machine.
As we know that in big-endian MSB Byte will store first. It means the MSB Byte will store at the lowest memory address. See the table,. In the little endian machine, LSB byte will store first. So the LSB Byte will store at the lowest memory address. Note: Some processor has the ability to switch one endianness to other endianness using the software that means it can perform like both big endian or little endian at a time.
This processor is known as the Bi-endian. As we know that in little endian machine least significant byte of any multibyte data field is stored at the lowest memory address.
So in the below program, we are checking the value of the lowest address. If the value is 1 then it will little endian either it will big endian. If your machine is little endian, the data in the memory will be something like the below expression:. The htons host-to-network-short and htonl host-to-network-long functions convert bit and bit values respectively from host machine to network byte order; the ntohs and ntohl functions convert from network to host byte order.
You can read more on endianness and information representation here. We can also write a small function to determine if an underlying machine architecture is little endian or big endian.
Hope you have enjoyed reading C program to check little and big endian architecture. Thanks for reading! So you can see that the actual size is 8 bytes. But, you may be wondering where the 3 extra bytes came from?
Lets come back to this later. We talked about data storage formats in the last section. For example, a char value is byte-aligned, which means that it can be stored at any address. A short value is 2 byte aligned, which means that it is always stored at an address which is a multiple of two. On the similar lines, integer and double values are 4 byte and 8 byte aligned respectively. Now, one might ask, how alignment helps? See, architecture of a computer processor is such that either it can read 4 bytes or 8 bytes in one go I am talking about 32 bit and 64 bit processors here.
So, if we talk about a 32 bit processor, it will be easy for it to read information packaged into 4 bytes. If an integer doesn't start at an address which is multiple of 4, it will cause performance issues as processor will have to do multiple cycles to read the integer value. Coming back to the example we discussed earlier. The reason those 3 extra bytes were added because compiler wanted to align integer to a 4 byte memory.
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