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使用如下api函数
bool getdiskfreespace(
lpctstr lprootpathname, // address of root path
lpdword lpsectorspercluster, // address of sectors per cluster
lpdword lpbytespersector, // address of bytes per sector
lpdword lpnumberoffreeclusters, // address of number of free clusters
lpdword lptotalnumberofclusters // address of total number of clusters
);
parameters
lprootpathname
points to a null-terminated string that specifies the root directory of the disk to return information about. if lprootpathname
is null, the function uses the root of the current directory.
lpsectorspercluster
points to a variable for the number of sectors per cluster.
lpbytespersector
points to a variable for the number of bytes per sector.
lpnumberoffreeclusters
points to a variable for the total number of free clusters on the disk.
lptotalnumberofclusters
points to a variable for the total number of clusters on the disk.
return values
if the function succeeds, the return value is nonzero.
if the function fails, the return value is zero. to get extended error information, call getlasterror.
例子：
procedure tform1.button1click(sender: tobject);
var drivestring:string;
sec1, byt1, cl1, cl2: longword;
disk_freespace : real;begin
getdiskfreespace(d:\, sec1, byt1, cl1, cl2);
disk_freespace := (cl1 / (1024*1024*1024))*sec1*byt1;
showmessage(format(该驱动器容量是%0.3fg,[disk_freespace]));
end;
上面的程序是将数据从字节单位转换为g的，之所以这样做，是为了避免当磁盘容量大于delphi基本数据类型所能存储的最大值，避免溢出。如果想获得以字节为单位的，那么将遇到大数相乘的问题。
下面提供一个大数相乘的算法，他接收两个字符串，输出这个两个字符串的乘积（当然字符串里都是数字）
function tform1.xaddy(x, y: string): string;
var
a,b,c:array[1..1000] of integer;
i,j,k,l,m,code:integer;
s,p,r:string;
begin
s := x; //两个要相乘的字符串
p := y;
l:=length(s);
for i:=l downto 1 do
val(s[i],a[l-i+1],code);
m:=length(p);
for i:=m downto 1 do
val(p[i],b[m-i+1],code);
for j:=1 to m do
for i:=1 to l do
begin
if c[i+j-1]+a[i]*b[j]<=9 then begin
c[j+i-1]:=c[i+j-1]+a[i]*b[j];
k:=i+j-1;
end else begin
c[j+i-1]:=c[i+j-1]+(a[i]*b[j]) mod 10;
c[j+i]:=c[j+i]+ c[j+i-1] div 10+ (a[i]*b[j]) div 10;
c[i+j-1]:=c[i+j-1] mod 10;
k:=i+j;
end;
end;
r := ;
for i:=k downto 1 do
r := r+inttostr(c[i]);
result := r;end;
下面我们就可以通过使用大数相乘的算法的得到磁盘的容量（用字节表示）
procedure tform1.button2click(sender: tobject);
var
sec1, byt1, cl1, cl2: longword;
disk_freespace : string;begin
getdiskfreespace(d:\, sec1, byt1, cl1, cl2);
disk_freespace := xaddy(inttostr(cl1),inttostr(sec1*byt1));
showmessage(format(该驱动器容量是%s字节,[disk_freespace]));
end;
程序完整的代码如下：
unit unit1;
interface
uses
windows, messages, sysutils, variants, classes, graphics, controls, forms,
dialogs, stdctrls;type
tform1 = class(tform)
button1: tbutton;
label1: tlabel;
label2: tlabel;
button2: tbutton;
label3: tlabel;
procedure button1click(sender: tobject);
procedure button2click(sender: tobject);
private
{ private declarations }
public
function xaddy(x : string;y:string) : string;
end;var
form1: tform1;implementation
{$r *.dfm}
procedure tform1.button1click(sender: tobject);
var drivestring:string;
sec1, byt1, cl1, cl2: longword;
disk_freespace : real;begin
getdiskfreespace(d:\, sec1, byt1, cl1, cl2);
disk_freespace := (cl1 / (1024*1024*1024))*sec1*byt1;
showmessage(format(该驱动器容量是%0.3fg,[disk_freespace]));
end;function tform1.xaddy(x, y: string): string;
var
a,b,c:array[1..1000] of integer;
i,j,k,l,m,code:integer;
s,p,r:string;
begin
s := x; //两个要相乘的字符串
p := y;
l:=length(s);
for i:=l downto 1 do
val(s[i],a[l-i+1],code);
m:=length(p);
for i:=m downto 1 do
val(p[i],b[m-i+1],code);
for j:=1 to m do
for i:=1 to l do
begin
if c[i+j-1]+a[i]*b[j]<=9 then begin
c[j+i-1]:=c[i+j-1]+a[i]*b[j];
k:=i+j-1;
end else begin
c[j+i-1]:=c[i+j-1]+(a[i]*b[j]) mod 10;
c[j+i]:=c[j+i]+ c[j+i-1] div 10+ (a[i]*b[j]) div 10;
c[i+j-1]:=c[i+j-1] mod 10;
k:=i+j;
end;
end;
r := ;
for i:=k downto 1 do
r := r+inttostr(c[i]);
result := r;end;
procedure tform1.button2click(sender: tobject);
var
sec1, byt1, cl1, cl2: longword;
disk_freespace : string;begin
getdiskfreespace(d:\, sec1, byt1, cl1, cl2);
disk_freespace := xaddy(inttostr(cl1),inttostr(sec1*byt1));
showmessage(format(该驱动器容量是%s字节,[disk_freespace]));
end;end.
