pro test
compile_opt idl2
a = 100.000000002
b = 100.000000000
help, a
help, b
if a gt b then $
print, '100.000000002 gt 100.000000000'
if a eq b then $
print, '100.000000002 eq 100.000000000'
if a lt b then $
print, '100.000000002 lt 100.000000000'
a = 100.000000002d
b = 100.000000000d
help, a
help, b
if a gt b then $
print, '100.000000002d gt 100.000000000d'
if a eq b then $
print, '100.000000002d eq 100.000000000d'
if a lt b then $
print, '100.000000002d lt 100.000000000d'
print, ''
a = double(100.000000002)
b = double(100.000000000)
help, a
help, b
if a gt b then $
print, 'double(100.000000002) gt double(100.000000000)'
if a eq b then $
print, 'double(100.000000002) eq double(100.000000000)'
if a lt b then $
print, 'double(100.000000002) lt double(100.000000000)'
print, ''
a = 100.000000002*1.0d
b = 100.000000000*1.0d
help, a
help, b
if a gt b then $
print, '100.000000002*1.0d gt 100.000000000*1.0d'
if a eq b then $
print, '100.000000002*1.0d eq 100.000000000*1.0d'
if a lt b then $
print, '100.000000002*1.0d lt 100.000000000*1.0d'
print, ''
a = double('100.1234567892')
b = double('100.1234567890')
help, a
help, b
if a gt b then $
print, '100.1234567892*1.0d gt 100.1234567890*1.0d'
if a eq b then $
print, '100.1234567892*1.0d eq 100.1234567890*1.0d'
if a lt b then $
print, '100.1234567892*1.0d lt 100.1234567890*1.0d'
print, ''
end
运行结果
A FLOAT = 100.000
B FLOAT = 100.000
100.000000002 eq 100.000000000
A DOUBLE = 100.00000
B DOUBLE = 100.00000
100.000000002d gt 100.000000000d
A DOUBLE = 100.00000
B DOUBLE = 100.00000
double(100.000000002) eq double(100.000000000)
A DOUBLE = 100.00000
B DOUBLE = 100.00000
100.000000002*1.0d eq 100.000000000*1.0d
A DOUBLE = 100.12346
B DOUBLE = 100.12346
str100.1234567892 gt str100.1234567890
初始的数据如果不是double型的，后面转为double型是没有用的，因为精度已经丢失
从a=100.000000002 可知，虽然是小数点后9位， 但IDL认为是float型
从a=100.000000002d 可知，加上d之后，IDL认为是double型
从a = double('100.1234567892') 可知，读入一个位数很多的字符串，将其转为double，可以最大限度的保持后续计算的精度