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Rational R1000/400

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⟦67da0e6f3⟧ Ada Source

    Length: 3072 (0xc00)
    Types: Ada Source
    Notes: 03_class, FILE, R1k_Segment, e3_tag, procedure Test_Binary_Tree, seg_0389f6

Derivation

└─⟦8527c1e9b⟧ Bits:30000544 8mm tape, Rational 1000, Arrival backup of disks in PAM's R1000
    └─ ⟦5a81ac88f⟧ »Space Info Vol 1« 
        └─⟦this⟧

E3 Source Code



with Text_Io;
with Binary_Tree;
procedure Test_Binary_Tree is
    N1, N2, N3, N4 : Binary_Tree.Node;
    Col : Integer := 1;
begin
    Text_Io.Put_Line ("merde");
    N1 := new Binary_Tree.Node_Structure;
    Text_Io.Put_Line ("toto");
    N2 := new Binary_Tree.Node_Structure;
    N3 := new Binary_Tree.Node_Structure;
    N4 := new Binary_Tree.Node_Structure;
    Text_Io.Put_Line ("titi");
    N1.Left_Son := N2;
    N1.Right_Son := N3;
    N3.Left_Son := N4;
    N1.The_Type := Binary_Tree.Liste_Type;
    N2.The_Type := Binary_Tree.Si_Type;
    N3.The_Type := Binary_Tree.Sinon_Type;
    N4.The_Type := Binary_Tree.Selon_Type;
    N1.S := new String'("salut...");
    N4.S := new String'("go...");
    Text_Io.Put_Line ("tutu");
    Binary_Tree.Put_Node (N1);
end Test_Binary_Tree;

E3 Meta Data

    nblk1=2
    nid=2
    hdr6=2
        [0x00] rec0=1b rec1=00 rec2=01 rec3=000
        [0x01] rec0=03 rec1=00 rec2=02 rec3=001
    tail 0x21733f6e084e08f918b81 0x42a00088462060003
Free Block Chain:
  0x2: 0000  00 00 00 1f 80 01 3b 01 00 15 65 6e 64 20 54 65  ┆      ;   end Te┆