In c ++, the usage of new is flexible. Here is a simple summary: "') X5 t # d/s # V) V9 N w! J
1. new () allocates a memory space of this type, and initializes this variable with the value in parentheses; 2. new [] allocates n Memory Spaces of this type and uses the default constructor to initialize these variables; # include <iostream>
# Include <cstring>
Using namespace std;) @ 8 B 'j-I, F W $ q $ Z
Int main () {7 o8 Z "h1 \ 4 J8 A8 R
// Char * p = new char ("Hello"); 6 @ 0 @ 4 A8 C % \ 6 R
// Error allocates a char (1 byte) space,
// Use "Hello" for initialization. This is obviously incorrect.
Char * p = new char [6];
// P = "Hello ";
// The character string cannot be directly assigned to the pointer p, because: 'c3 {6 _) l7 P % Q2 u "S
// The pointer p points to the first character of the string and can only use the following * I3 m * Z2 d; A + \ "L7 m # t
// Strcpy
Strcpy (p, "Hello ");
Cout <* p <endl; // only outputs the first character of the string pointed to by p!
Cout <p <endl; // output the string pointed to by p!
Delete [] p;-q % k E w; Y
Return 0;} 7 U) m4 | * B8 e 'P: v
Output result:
H
Hello; P0 X % Z! [5 V8 j; l! ~; ^ & D /\
3. When the new operator is used to define a multi-dimensional array variable or array object, it generates a pointer to the first element of the array. The returned type retains all dimensions except the leftmost dimension. For example:
Int * p1 = new int [10];
Returns an int x + l $ U; a + [0 U8 K
Int (* p2) [10] = new int [2] [10];
A two-dimensional array is added, and the leftmost one [2] is removed, with int [10] Left. Therefore, a pointer int (*) pointing to a one-dimensional array such as int [10] is returned (*) [10].
Int (* p3) [2] [10] = new int [5] [2] [10]; new a three-dimensional array, remove the leftmost one [5], there is also int [2] [10], so a pointer to the two-dimensional array int [2] [10] type int (*) [2] [10] is returned.
# Include <iostream>
# Include <typeinfo>
Using namespace std;-@. Q B $ f: [(h
Int main (){
Int * a = new int [34];
Int * B = new int [];
Int (* c) [2] = new; M/L1] 3 L) I] "G1 s" x/s & H
Int [34] [2];
Int (* d) [2] = new int [] [2];
Int (* e) [2] [3] = new int [34] [2] [3];) s, _ 6 B & f4 j/C-z/Y/L-u, g "p
Int (* f) [2] [3] = new int [] [2] [3];
A [0] = 1; + p6 N0 A "v" W: j3 e
B [0] = 1; // runtime error, no memory allocation, B only acts as a pointer, used to point to the corresponding data/P9 N _ + _-R .\
C [0] [0] = 1;
D [0] [0] = 1; // runtime error, no allocated memory, d only acts as a pointer, used to point to the corresponding data-i0 ^: D (C; n-! V "J
E [0] [0] [0] = 1;
F [0] [0] [0] = 1; // runtime error, no memory allocated. f only acts as a pointer and is used to point to the corresponding data
Cout <typeid (a). name () <endl ;! U. d "d # q (I % R
Cout <typeid (B). name () <endl;: B8 L % z-W4 V6 e $ y3 y! [& D
Cout <typeid (c). name () <endl;
Cout <typeid (d). name () <endl;. _ 9 h9 K3 g /?
Cout <typeid (e). name () <endl;
Cout <typeid (f). name () <endl;. j9' % R * S) l * a8 u "}; D *~
Delete [] a; delete [] B; delete [] c;
Delete [] d; delete [] e; delete [] f; 5] 5 h0 S "F3 g-r-d. b8 R/j
} & Y6 r1 l0 j & I % r6 f & C1
Output result:
Int * (} "q #]" u % v) P! \ 8 _ "Y! ]
Int *
Int (*) [2]
Int (*) [2]
Int (*) [2] [3] $ v/u (a: p &[
Int (*) [2] [3]
Note the differences between parentheses () and brackets []. Otherwise, errors may occur. I have suffered a loss on this issue. I hope you will pay more attention to it.
