var ts:array[1..20,1..20] of integer;
ns:array[1..20,1..20] of integer;procedure useuse(ts:array of integer;var ns:array of integer);
begin
............
end;
调试通不过,将ts,ns改为一维数组又可以,为什么啊?该怎样调用?
谢谢了,在线等待!!!!!!!!!!!1
ns:array[1..20,1..20] of integer;procedure useuse(ts:array of integer;var ns:array of integer);
begin
............
end;
调试通不过,将ts,ns改为一维数组又可以,为什么啊?该怎样调用?
谢谢了,在线等待!!!!!!!!!!!1
begin
............
end;
begin
............
end;
不行啊!!!!!!!!
var ts:array[1..20,1..20] of integer;
ns:array[1..20,1..20] of integer;procedure useuse(ts:array of integer;var ns:array of integer);
begin
............
end;
调用的时候这样写
useuse(ts[0], ns[0]);
begin
............
end;这样呢?
ns:array[1..20,1..20] of integer;procedure useuse(ts:array[1..20,1..20] of integer;var ns:array[1..20,1..20] of integer);
begin
............
end
type
tarray = array[1..20,1..20] of integer;
var
ts, na: tarray;
procedure useuse(ts, ns: tarray);
begin
............
end;
楼主,如果你要在过程中设置二维数组的所有值,你可以这样做
还是用你那个二维数组
procedure use(a :array of integer;b:array of integer;size1, size2 :integer);
var
i :integer;
begin
for i :=0 to size1 do
a[i] :=0;
for i := 0 to size2 do
b[i] :=0;
end;
procedure use(var a, b :array of integer; size1, size2 :integer);
TDoubleArrSingle = array[0..2, 0..2] of Single;function GetCont(var CombH: TDoubleArrSingle; var CombHT:TDoubleArrSingle): Boolean;
var ts:array[1..20,1..20] of integer;
ns:array[1..20,1..20] of integer;procedure useuse(ts:array[1..20,1..20] of integer;var ns:array[1..20,1..20] of integer);
begin
............
end
这个方法是不可以的,请使用我提到的第二个方法,即声明一个二维数组类型