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/420023/10822923
txt 2dfft.txt
#include
#include
#include
#include
#define pi 3.141592654
struct COMPLEX{
float REAL;
float IMAGE;
};
void Bit_Reversal(unsigned int *,int ,int
www.eeworm.com/read/419929/10828428
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/275048/10837747
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/275048/10837767
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/275044/10837829
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/275022/10838674
c 7ngin.c
#include "math.h"
#include "stdlib.h"
#include "6gmiv.c"
int ngin(m,n,eps1,eps2,x,ka,f,s)
void (*f)(),(*s)();
int m,n,ka;
double eps1,eps2,x[];
{ int i,j,k,l,kk,jt;
doubl
www.eeworm.com/read/275022/10838701
c 7netn.c
#include "stdlib.h"
#include "math.h"
#include "stdio.h"
#include "6gaus.c"
int netn(n,eps,t,h,x,k,f)
int n,k;
void (*f)();
double eps,t,h,x[];
{ int i,j,l;
double am,
www.eeworm.com/read/275022/10838863
c 12lplq.c
#include "stdlib.h"
#include "4rinv.c"
#include "4trmul.c"
int lplq(a,b,c,m,n,x)
int m,n;
double a[],b[],c[],x[];
{ int i,mn,k,j,l,*js;
double s,z,dd,y,*p,*d;
js=mall
www.eeworm.com/read/275022/10838998
c 6gmqr.c
#include "4maqr.c"
#include "stdlib.h"
int gmqr(a,m,n,b,q)
int m,n;
double a[],b[],q[];
{ int i,j;
double d,*c;
int maqr(double [],int,int,double []);
c=malloc(n*size
www.eeworm.com/read/275022/10839008
c 6grad.c
#include "4trmul.c"
#include "math.h"
#include "stdlib.h"
#include "stdio.h"
void grad(a,n,b,eps,x)
int n;
double a[],b[],x[],eps;
{ int i,k;
double *p,*r,*s,*q,alpha,bet