代码搜索：malloc
找到约 10,000 项符合「malloc」的源代码
代码结果 10,000
www.eeworm.com/read/349004/10856292
c dfpopt.c
#include "hjfgf.c"
#include "stdlib.h"
void gradient(double x[],double g[],int n)
{int i;
double af,f1,f2,dltx=0.000001;
for(i=0;i
www.eeworm.com/read/349004/10856305
c powell.c
#include "hjfgf.c"
double oneoptim(double x0[],double s[],double h0,double epsg,int n,double x[])
{double *a,*b,ff;
a=(double *)malloc(n*sizeof(double));
b=(double *)malloc(n*sizeof(double));
www.eeworm.com/read/348948/10859254
c conversions.c
/*
*
* GWIC
*
* (c) Joonas Lehtinen (jole@jole.fi), TUCS, 1998
*
* modification for rgb2yuv & yuv2rgb
* for color model YUV422 by Sasha Chukov
*
*/
#include "gwic.h"
/* Convert and round fl
www.eeworm.com/read/274540/10866387
txt 1.txt
#include
#include
#include
#include
#define N 50
#define M 500
char *a[32]={"auto","break","case","char","const","continue","default","do",
www.eeworm.com/read/274170/10886390
c dynamic.c
# include
# include
# define NUM 10
int main()
{
char *str[NUM]; /* 定义一个字符性的指针数组 */
int t;
/* 为数组中的每个指针分配内存 */
for(t=0; t
www.eeworm.com/read/274170/10886454
c euler.c
#include "stdio.h"
#include "stdlib.h"
#include
int Func(y,d)
double y[],d[];
{
d[0]=y[1]; /*y0'=y1*/
d[1]=-y[0]; /*y1'=y0*/
d[2]=-y[2]; /*y2'=y2*/
return(1);
}
void E
www.eeworm.com/read/274170/10886468
c grkt10.c
#include "stdio.h"
#include "stdlib.h"
void RKT(t,y,n,h,k,z)
int n; /*微分方程组中方程的个数，也是未知函数的个数*/
int k; /*积分的步数(包括起始点这一步)*/
double t; /*积分的起始点t0*/
double h; /*积分的步长*/
double y[]; /
www.eeworm.com/read/273950/10894094
cpp powell.cpp
/*本程序包含5个C文件: mpowell.c, powell.c,
funct.c(目标函数), jtf.c(进退法), hjfgf.c(黄金分割法)*/
//题目 y=x1*x1+x2*x2-x1*x2-10*x1-4*x2+60
//#include "powell.c"
#include "stdio.h"
#include "stdlib.h"
#include "ma
www.eeworm.com/read/273736/10903028
cpp extra2.cpp
/*#include
#include
#include
#define MAX(x,y) (x>y?x:y)
int LCS(char *c1,char *c2){
int len1=strlen(c1),len2=strlen(c2),i,j;
int **matrix=(int**)malloc(sizeof(int)*(
www.eeworm.com/read/273564/10910833
cpp tanxin.cpp
#include
#include
#include
int MAX(int *D,int i, int j);
int FIND(int *parent,int i);
int MIN(int n,int m) ;
int FJS(int *D,int n,int