Swift 中的指针和 malloc

Question

我试图转换这为迅速。

内存分配逻辑面临的问题

Byte *p[10000];

p[allocatedMB] = malloc(1048576);
memset(p[allocatedMB], 0, 1048576);

如何快速写这个？

Answer 1

您可以malloc从 Swift使用，它返回一个“原始指针”：

var p: [UnsafeMutableRawPointer?] = Array(repeating: nil, count: 10000)
var allocatedMB = 0

p[allocatedMB] = malloc(1048576)
memset(p[allocatedMB], 0, 1048576)

或者，使用UnsafeMutablePointer及其 allocate和initialize方法：

var p: [UnsafeMutablePointer<UInt8>?] = Array(repeating: nil, count: 10000)
var allocatedMB = 0

p[allocatedMB] = UnsafeMutablePointer.allocate(capacity: 1048576)
p[allocatedMB]?.initialize(to: 0, count: 1048576)