习题6.5

#include <stdio.h>
#include <stdlib.h>

int main()
{
float c,f;
for (f=0;f<=300;f=f+10)
{
c=5.0/9*(f-32);
printf ("%f    %f\n",f,c);
}
return 0;
}

结果如下：

c,f

0.000000 -17.777779

10.000000 -12.222222

20.000000 -6.666667

30.000000 -1.111111

40.000000 4.444445

50.000000 10.000000

60.000000 15.555555

70.000000 21.111111

80.000000 26.666666

90.000000 32.222221

100.000000 37.777779

110.000000 43.333332

120.000000 48.888889

130.000000 54.444443

140.000000 60.000000

150.000000 65.555557

160.000000 71.111115

170.000000 76.666664

180.000000 82.222221

190.000000 87.777779

200.000000 93.333336

210.000000 98.888885

220.000000 104.444443

230.000000 110.000000

240.000000 115.555557

250.000000 121.111115

260.000000 126.666664

270.000000 132.222229

280.000000 137.777771

290.000000 143.333328

300.000000 148.888885

Process returned 0 (0x0) execution time : 0.020 s

Press any key to continue.

心得体会：

1.有float时小数保留六位，所以是5.0/9而不是5/9，否则结果输出c全为0.