std:: memcpy

Copies count bytes from the object pointed to by src to the object pointed to by dest . Both objects are reinterpreted as arrays of unsigned char .

If the objects overlap, the behavior is undefined.

If either dest or src is an invalid or null pointer, the behavior is undefined, even if count is zero.

If the objects are potentially-overlapping or not TriviallyCopyable , the behavior of memcpy is not specified and may be undefined.

Contents

[edit] Parameters

dest - pointer to the memory location to copy to
src - pointer to the memory location to copy from
count - number of bytes to copy

[edit] Return value

[edit] Notes

std::memcpy may be used to implicitly create objects in the destination buffer.

std::memcpy is meant to be the fastest library routine for memory-to-memory copy. It is usually more efficient than std::strcpy , which must scan the data it copies or std::memmove , which must take precautions to handle overlapping inputs.

Several C++ compilers transform suitable memory-copying loops to std::memcpy calls.

Where strict aliasing prohibits examining the same memory as values of two different types, std::memcpy may be used to convert the values.

[edit] Example

Run this code
#include #include #include int main() { // simple usage char source[] = "once upon a daydream. ", dest[4]; std::memcpy(dest, source, sizeof dest); std::cout  "dest[4] = ; for (int n{}; char c : dest) std::cout  (n++ ? ", " : "")  '\''  c  "'"; std::cout  ">;\n"; // reinterpreting double d = 0.1; // std::int64_t n = *reinterpret_cast(&d); // aliasing violation std::int64_t n; std::memcpy(&n, &d, sizeof d); // OK std::cout  std::hexfloat  d  " is "  std::hex  n  " as a std::int64_t\n"  std::dec; // object creation in destination buffer struct S { int x{42}; void print() const { std::cout  '  x  ">\n"; } } s; alignas(S) char buf[sizeof(S)]; S* ps = new (buf) S; // placement new std::memcpy(ps, &s, sizeof s); ps->print(); }
dest[4] = ; 0x1.999999999999ap-4 is 3fb999999999999a as a std::int64_t