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guyskk
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用 Python 批量转换 C 源码遇到问题

  •  1
     
  •   guyskk · 2017-01-12 13:57:54 +08:00 · 805 次点击
    这是一个创建于 1106 天前的主题,其中的信息可能已经有所发展或是发生改变。

    有 40 多个 C 源码文件,每一个里面都有 main 函数,都是从 stdin 或者文件读取参数,结果直接 print 到 stdout 或者 stderr 。 我想把它全部改写成普通函数,直接传参数调用函数,然后返回状态码(编译成动态链接库然后从 Python 里面调用)。 我尝试用正则替换实现,但是只能做到把 errorprint 函数提取出来,读取参数和输出不知道怎么处理。 求 V 友们指点,非常感谢!

    样本 1 :

    #include <stdio.h>
    #include <math.h>
    #include <stdlib.h>
    
    #define Pi 3.141592654
    
    
    double aa[4096];
    void error_print();
    
    void main(int argc, char *argv[]){
        double *x, dum;
        int i ,  nn,  j, im, jm ;
        long ik;
        FILE *fr ;
    
        if (argc< 2 || argc >2 ){
            error_print(argv[0]);
            exit(1);
        }
    
    
    
        fr=fopen(argv[1],"r");
        x = (double *) calloc( 1, sizeof(double) );
    
        ik=0;
        while ((fscanf(fr,"%le", &dum))!= EOF){
            x = (double *) realloc ( x, sizeof(double)*(ik+1));
    
            if(x == NULL ){
                fprintf(stderr,"%s: Allocation error while reading, exiting\n", argv[0]);
                exit(1);
            }
            x[ik]=dum;	 
    	/* fprintf(stderr," %e \n", x[ik]); */
            ik++;
        }
        nn=(int) sqrt( (double) ik);
       fclose(fr); 
    
    
        
        for(i=1; i<=nn; i++){
          im = nn+1-i;
          
            for (j=1; j<=nn; j++){
    	  double fdd=0;
    	  jm=nn+1-j;
    	
    	  ik=(i-1)*nn+j-1;
    	  fdd += x[ik];
    	  ik=(j-1)*nn+i-1;
    	  fdd += x[ik];	  
    	  ik=(j-1)*nn+im-1; 
    	  fdd += x[ik];
    	  ik=(i-1)*nn+jm-1; 
    	  fdd += x[ik];
    
    /*	  ik=(j-1)*nn+i-1;
    	  fdd += x[ik];
    	  ik=(j-1)*nn+im-1;  
    	  fdd += x[ik];	  */
    	    printf("%g\n", fdd);
    	    }
          printf("\n");
    }
    
    
        free(x);
    
        fclose(fr);
    }
    
    void error_print(char *arr) { 
    
        fprintf(stderr,"\n%s  converts the intensity file into sum of itself three times rotated 90 degrees\n", arr);
    
        fprintf(stderr,"\nUSAGE: %s F > M_F  where F is the file in gnuplot format, M_F is the output file in the gnuplot format\n\n", arr);
    
    }
    

    样本 2

     
    #include "pipes.h"
    #include <math.h>
    #include <string.h>
    
    void main(int argc, char *argv[]){
        void error_print();
    
        int i,j, n2;
        double sum, sum1r, sum1i, sum1, sum2, dx,dx2, s_p, x,y, x_c, y_c;
        double sum1x, sum1y;
        long ik1;
    
    
    
        /* Processing the command line argument  */
    
        if (argc!= 2){
            error_print(argv[0]);
            exit(1);
        }
    
    
        read_field();
    
        dx =field.size/(field.number);
        dx2 = dx*dx;
        n2=field.number/2+1;
    
    
    
        /* Calculating the power */
        sum=sum1r=sum1i=sum2=0.;
        ik1=0;
        for (i=1;i<=field.number ;i++){
            for (j=1;j<=field.number ;j++){
    
    
    
                
    	      s_p=(field.real[ik1]*field.real[ik1]+ \
    field.imaginary[ik1]*field.imaginary[ik1]);
    	    sum2 += s_p;
    	    sum += sqrt(s_p);
                sum1r += field.real[ik1];
                sum1i += field.imaginary[ik1];
                ik1++;
            }
        }
        sum1=(sum1r*sum1r+sum1i*sum1i);
    
    
        if (sum == 0) {
    
    fprintf(stderr,"Strehl: Zero beam power, program terminated\n");
            exit(1);
    			      }
    
    if(strstr(argv[1], "y")!= NULL)fprintf(stderr,"Strehl: ratio= %e energy= %e\n",sum1/sum/sum, sum2*dx2);
    
    /* Calculating the center of gravity: */
     sum=sum1r=sum1i=sum2=0.;
     ik1=0;
        for (i=1;i<=field.number ;i++){
          y=(i-n2)*dx;
            for (j=1;j<=field.number ;j++){
    	  x=(j-n2)*dx;
    	  sum2=(field.real[ik1]*field.real[ik1]\
    +field.imaginary[ik1]*field.imaginary[ik1]);
    	  sum1r += sum2*x;
    	  sum1i += sum2*y;
    	  sum += sum2;
    
                ik1++;
            }
        }
    
        x_c=sum1r/sum;
        y_c=sum1i/sum;
    
       fprintf(stderr,"Center_of_gravity: x= %e y= %e\n", x_c, y_c);
      
    
    /* Calculating moments of the distribution */
     sum1r=sum1x=sum1y=0.;
     ik1=0;
        for (i=1;i<=field.number ;i++){
          double y_y_c;
          y=(i-n2)*dx;
          y_y_c=y-y_c;
          
            for (j=1;j<=field.number ;j++){
    	  double temp_int, x_x_c;
    	  x=(j-n2)*dx;
    	  x_x_c=x-x_c;
    	  temp_int = (field.real[ik1]*field.real[ik1]\
    +field.imaginary[ik1]*field.imaginary[ik1]);
    	  sum1r += temp_int*(x_x_c*x_x_c+y_y_c*y_y_c);
    	  sum1x += temp_int*(x_x_c*x_x_c);
    	  sum1y += temp_int*(y_y_c*y_y_c);
    
                ik1++;
            }
        }
    
    
       fprintf(stderr,"Standard deviation:  S_r=%e S_x= %e S_y= %e\n", sqrt(sum1r/sum), sqrt(sum1x/sum), sqrt(sum1y/sum));
     fprintf(stderr,"Grid size: %e, Grid sampling: %d\n", field.size, field.number); 
    
    
        write_field();
    
    
    }
    
    
    void error_print(char *arr) { 
    
        fprintf(stderr,"\n%s: prints the general info to the stderr\n",arr);
    
    
        fprintf(stderr,"\n%s y, y prevents arrival of this message\n\n",arr);
    
    
    
    
    }
    
    1 回复  |  直到 2017-06-20 12:38:45 +08:00
    wentian
        1
    wentian   2017-06-20 12:38:45 +08:00
    要是我, 我会手动去做这件事情

    如果要用代码实现, 可能一天过去了, 一个屁都还没有写出来
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