How do I know what the natural alignment of a type should be?

Question

Ryan McCombe · Accepted Answer

Natural alignment is typically based on the size of the data type. Here's a practical guide to common alignment requirements:

1#include <iostream>
2
3int main() {
4  std::cout << "Alignment requirements:\n"
5    << "char: " << alignof(char) << " bytes\n"
6    << "short: " << alignof(short) << " bytes\n"
7    << "int: " << alignof(int) << " bytes\n"
8    << "long: " << alignof(long) << " bytes\n"
9    << "float: " << alignof(float) << " bytes\n"
10    << "double: " << alignof(double) << " bytes\n"
11    << "pointer: " << alignof(int*) << " bytes\n";
12}

1Alignment requirements:
2char: 1 bytes
3short: 2 bytes
4int: 4 bytes
5long: 8 bytes
6float: 4 bytes
7double: 8 bytes
8pointer: 8 bytes

For compound types like structs and classes, the alignment is typically determined by the largest alignment requirement of any member:

1#include <iostream>
2
3struct Example {
4  char A;// 1 byte alignment
5  double B;// 8 byte alignment
6  int C;// 4 byte alignment
7};
8
9int main() {
10  std::cout << "Struct alignment: "
11    << alignof(Example) << " bytes\n";
12}

1Struct alignment: 8 bytes

You can use the alignas specifier to request specific alignment:

1#include <iostream>
2
3struct alignas(16) Custom {
4  int Value;
5};
6
7int main() {
8  std::cout << "Custom alignment: "
9    << alignof(Custom) << " bytes\n";
10
11  // This will always start at a 16-byte boundary
12  Custom Instance;
13}

Remember:

Basic types are usually aligned to their size
Compound types align to their largest member
The alignof() operator tells you alignment requirements
Arrays align to their element type's requirements
Alignment is always a power of 2

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