代码搜索：malloc

找到约 10,000 项符合「malloc」的源代码

代码结果 10,000
热门代码： main.c GPIO I2C SPI UART timer interrupt ADC PWM LED
www.eeworm.com/read/431721/8659367

c 10adms.c

#include "stdlib.h" #include "math.h" void adms(t,h,n,y,eps,k,z,f) void (*f)(); int n,k; double t,h,eps,y[],z[]; { void rkt(); int i,j,m; double a,q,*b,*e,*s,*g,*d;
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www.eeworm.com/read/431721/8659369

c 10wity.c

#include "stdlib.h" void wity(t,y,n,h,k,z,f) void (*f)(); int n,k; double t,h,y[],z[]; { int i,j; double x,*a,*d; a=malloc(n*sizeof(double)); d=malloc(n*sizeof(doubl
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www.eeworm.com/read/431721/8659373

c 10gil.c

#include "stdlib.h" #include "math.h" void gil(t,h,y,n,eps,q,f) void (*f)(); int n; double t,h,eps,y[],q[]; { int i,j,k,m,ii; double x,p,hh,r,s,t0,dt,qq,*d,*u,*v,*g; s
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www.eeworm.com/read/431721/8659557

c 4ssgj.c

#include "stdlib.h" #include "math.h" #include "stdio.h" int ssgj(a,n) int n; double a[]; { int i,j,k,m; double w,g,*b; b=malloc(n*sizeof(double)); for (k=0; k
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c 9mtml.c

#include "stdlib.h" #include "math.h" #include "3rnd1.c" double mtml(n,a,b,f) int n; double a[],b[],(*f)(); { int m,i; double r,s,d,*x; x=malloc(n*sizeof(double));
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www.eeworm.com/read/431721/8659716

c 15topo.c

#include "stdlib.h" void topo(n,r,m,p) int n,m,p[],r[]; { int top,i,j,k,t,*s,*g,*f; struct node { int suc; int next; } *q; q=(struct node *)malloc(m*sizeo
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c 13lman.c

#include "stdlib.h" #include "4rinv.c" int lman(n,m,k,f,q,r,h,y,x,p,g) int n,m,k; double f[],q[],r[],h[],y[],x[],p[],g[]; { int i,j,kk,ii,l,jj,js; double *e,*a,*b; e=mallo
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www.eeworm.com/read/287904/8662667

c algo6-1.c

/* algo6-1.c 求赫夫曼编码。实现算法6.12的程序 */ #include"c1.h" #include"c6-7.h" int min1(HuffmanTree t,int i) { /* 函数void select()调用 */ int j,flag; unsigned int k=UINT_MAX; /* 取k为不小于可能的值 */
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www.eeworm.com/read/431653/8663974

cpp 哈夫曼树.cpp

#include #include #include #include #include //typedef int TElemType; const int UINT_MAX=1000; typedef struct { int weight;
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www.eeworm.com/read/387456/8675527

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
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