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【求助】程序運(yùn)行已經(jīng)無錯(cuò)誤,但是輸入文件有問題,求教。
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輸入文件為: 0.52 5.28794 8.0 13.8400 17.95 20.2000 28.65 24.9000 50.65 31.1000 104.6 36.5000 156.6 36.6000 260.7 31.0000 364.4 20.9000 468.0 7.80000 507.0 1.50000 520.0 0.200000 1.86548 -0.046115 4.0 14.0 30.0 60.0 130.0 230.0 450.0 515.0 結(jié)果出現(xiàn)這個(gè)問題:[Session started at 2011-04-06 17:46:04 +0800.] GNU gdb 6.3.50-20050815 (Apple version gdb-1469) (Wed May 5 04:36:56 UTC 2010) Copyright 2004 Free Software Foundation, Inc. GDB is free software, covered by the GNU General Public License, and you are welcome to change it and/or distribute copies of it under certain conditions. Type "show copying" to see the conditions. There is absolutely no warranty for GDB. Type "show warranty" for details. This GDB was configured as "x86_64-apple-darwin".tty /dev/ttys000 Loading program into debugger… Program loaded. run [Switching to process 1601] Running… 輸入文件名:splin1.txt 不能打開這個(gè)文件splin1.txt Debugger stopped. Program exited with status value:1.(gdb) |
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我用VC6運(yùn)行了你的程序,采用的也是你給出的例子文件,沒有問題,可以輸出結(jié)果。如下: 0.52 5.28794 -6.27744e+066 2.84681e+066 8 13.84 1.90782e+066 -6.58235e+065 17.95 20.2 -5.40918e+065 1.66027e+065 28.65 24.9 1.93591e+065 -2.87356e+064 50.65 31.1 -7.10913e+064 4.6735e+063 104.6 36.5 1.61149e+064 -1.44065e+063 156.6 36.6 5.22698e+063 1.02188e+063 260.7 31 -6.36429e+064 -2.34503e+063 364.4 20.9 2.48875e+065 8.37238e+063 468 7.8 -9.3144e+065 -3.11584e+064 507 1.5 2.47047e+066 2.05615e+065 520 0.2 -6.27744e+066 -2.89728e+066 -6.27744e+066 1.93191e+265 -9.23265e+198 2.94154e+132 -6.27744e+066 1.93191e+265 -9.23265e+198 2.94154e+132 -6.27744e+066 1.93191e+265 -9.23265e+198 2.94154e+132 -6.27744e+066 1.93191e+265 -9.23265e+198 2.94154e+132 -6.27744e+066 1.93191e+265 -9.23265e+198 2.94154e+132 -6.27744e+066 1.93191e+265 -9.23265e+198 2.94154e+132 -6.27744e+066 1.93191e+265 -9.23265e+198 2.94154e+132 -6.27744e+066 1.93191e+265 -9.23265e+198 2.94154e+132 6.20039e+068 你的問題是不是由于你的輸入文件還在打開著?在執(zhí)行程序的時(shí)候,要關(guān)閉它 |
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void splin1::interp() { int i,j; double h0,h1,alpha,beta; s[0]=dy[0];dy[0]=0.0; h0=x[1]-x[0]; for(j=1;j<=n-2;j++) { h1=x[j+1]-x[j]; alpha=h0/(h0+h1); beta=(1.0-alpha)*(y[j]-y[j-1])/h0; beta=3.0*(beta+alpha*(y[j+1]-y[j])/h1); dy[j]=-alpha/(2.0+(1.0-alpha)*dy[j-1]); s[j]=(beta-(1.0-alpha)*s[j-1]); s[j]=s[j]/(2.0+(1.0-alpha)*dy[j-1]); h0=h1; } for(j=n-2;j>=0;j--) dy[j]=dy[j]*dy[j+1]+s[j]; for(j=0;j<=n-2;j++)s[j]=x[j+1]-x[j]; for(j=0;j<=n-2;j++) { h1=s[j]*s[j]; ddy[j]=6.0*(y[j+1]-y[j])/h1-2.0*(2.0*dy[j]+dy[j+1])/s[j]; } h1=s[n-2]*s[n-2]; ddy[n-1]=6.*(y[n-2]-y[n-1])/h1+2.*(2.*dy[n-1]+dy[n-1])/s[n-2]; integ=0.0; for(i=0;i<=n-2;i++) { h1=0.5*s*(y+y[i+1]); h1=h1-s*s*s*(ddy+ddy[i+1])/24.0; integ=integ+h1; } for(j=0;j<=m-1;j++) { if(t[j]>=x[n-1])i=n-2; else { i=0; while(t[j]>x[j+1])i=i+1; } h1=(x[j+1]-t[j])/s; h0=h1*h1; z[j]=(3.0*h0-2.0*h0*h1)*y; z[j]=z[j]+s*(h0-h0*h1)*dy; dz[j]=6.0*(h0-h1)*y/s; dz[j]=dz[j]+(3.0*h0-2.0*h1)*dy; ddz[j]=(6.0-12.0*h1)*y/(s*s); ddz[j]=ddz[j]+(2.0-6.0*h1)*dy/s; h1=(t[j]-x)/s; h0=h1*h1; z[j]=z[j]+(3.0*h0-2.0*h0*h1)*y[i+1]; z[j]=z[j]-s*(h0-h0*h1)*dy[i+1]; dz[j]=dz[j]-6.0*(h0-h1)*y[i+1]/s; dz[j]=dz[j]+(3.0*h0-2.0*h1)*dy[i+1]; ddz[j]=ddz[j]+(6.0-12.0*h1)*y[i+1]/(s*s); ddz[j]=ddz[j]-(2.0-6.0*h1)*dy[i+1]/s; } } 我就是這一段讀不過去。。。 |
金蟲 (著名寫手)
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看來樓主對于path這種東西沒概念... VC里面默認(rèn)的“當(dāng)前工作路徑”使得你在VC里面點(diǎn)擊那個(gè)小三角符號運(yùn)行程序時(shí),默認(rèn)的路徑就是程序文件所在的那個(gè)路徑,所以這時(shí)候你直接輸入a.txt,它能找到 而xcode設(shè)定的當(dāng)前工作路徑和VC的習(xí)慣可能不一樣(從你的例子來看是應(yīng)該不一樣),總之你可以先輸入a.txt的絕對路徑嘗試一下 至于怎么在xcode里面設(shè)定工作路徑,這個(gè)樓主慢慢找吧... |
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