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program 9
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C ifの使い方を知るためだけの、殆ど無意味なプログラム 
C  (何をやってるか、考えてみてください) */
      program main
      real*8 x

      do i=1,10
        read(*,*)x
        if( x.gt.10)then
            write(*,*)x," > 10"
        else if(x.gt.5)then
            write(*,*)"10 >= ",x," > 5"
        else if(x.gt.0)then
            write(*,*)"5 >= ",x," > 0"
        else
            write(*,*)x," <= 0 "
            stop
        endif
      enddo
      end
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program 10
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C if文の利用例として2分法で非線形方程式の解を求める。
C 下の例題では
C x^3 + 5x^2 - 3x -5 = 0
C の解を求めている(^はべき乗を表す)。

      program main
      parameter(nmax=50)
      real*8 x1,x2,x3   
      real*8 f1,f2,f3   

C 解探索の範囲を指定
      read(*,*)x1,x2    

C 2分法を最大nmax回くりかえす
      do irepeat = 1,nmax
        f1 = x1**3 + 5*x1**2 - 3*x1 -5 
        f2 = x2**3 + 5*x2**2 - 3*x2 -5 

C  x1とx2の差が与えた精度以内なら終了
        if(abs((x1-x2)/x1).le. 1e-10)then 
          write(*,*)"convergence"      
          write(*,*)"x1 = ",x1," f1 = ",f1
          write(*,*)"x2 = ",x2," f2 = ",f2
          stop                            

C  収束してないと判断されたとき。
C  f1とf2の符号が逆なら範囲を二分する
        else if(f1*f2.lt.0)then           
            x3 = (x1+x2)/2                
            f3 = x3**3 + 5*x3**2 - 3*x3 -5
            if(f1*f3.lt.0)then            
              x2=x3                      
            else                         

              x1=x3                      

            endif                    
       write(*,*)irepeat," step  x1=",x1,"  f1=",f1,"  x2=",x2,"  f2=",f2 

C  f1とf2の積が正なら、この範囲に解がないと判断して終了
        else
          write(*,*)"no solution is found between ",x1," and ",x2
          stop                            
        endif                           
      enddo                             

C  nmax回のくりかえしで解に収束しなかったときは終了
      write(*,*)"not converged after ",nmax," steps"
      end