代码搜索:fprintf
找到约 10,000 项符合「fprintf」的源代码
代码结果 10,000
www.eeworm.com/read/305833/13760076
c pstasks.c
#include
#include
/*#include "store.h"*/
#include "input.tab.h"
#include "crono.h"
#define MAX_TIME 550
#define X_PUNTOS 500 /*Puntos PostScript eje X*/
#define Y_PUNTOS 650 /*Pu
www.eeworm.com/read/305704/13762923
cpp 1.cpp
/*
C语言词法分析程序
4.25
*/
#include
#include
#include "2.h"
#define KEYNUM 32
//output form
struct output
{
int type;
int value;
int lineno;
//int column;
};
//keywords
www.eeworm.com/read/305243/13776228
m pcamat.m
function [E, D] = pcamat(vectors, firstEig, lastEig, s_interactive, ...
s_verbose);
%PCAMAT - Calculates the pca for data
%
% [E, D] = pcamat(vectors, firstEig, lastEig, ...
% int
www.eeworm.com/read/305004/13780431
h alsactl.h
extern int debugflag;
extern int force_restore;
extern char *command;
#if __GNUC__ > 2 || (__GNUC__ == 2 && __GNUC_MINOR__ >= 95)
#define error(...) do {\
fprintf(stderr, "%s: %s:%d: ", command, __F
www.eeworm.com/read/304826/13786140
txt 10-07.txt
%例10-7 只有一种选择的情况下if-else-end选择语句的使用。
%该函数用于演示if-else-end语句的第一种用法
function f=ifone(x)
if x>0
fprintf('%f is a positive number\n',x)
end
www.eeworm.com/read/304826/13786175
txt 10-08.txt
%例10-8 有两种选择的情况下if-else-end选择语句的使用。
%该程序用于演示有2种选择时if-else-end语句的使用
function iftwo(x)
if x>0
fprintf('%f is a positive number\n',x)
else
fprintf('%f is not a positive number\n',x)
en
www.eeworm.com/read/304789/13787058
c unionlog.c
#include
#include
#include
#include
#ifdef _WIN32
# include
# include
#else
#ifdef _UNIX
# include
struct UnionTM
{
www.eeworm.com/read/304730/13788030
cpp main.cpp
#include "Cubic.cpp"
#include
#include
void main()
{
int i,j,p,m,mSeqLength,sequence[1000];
int length;
int num1,num2,num3;
//complex complex1;
www.eeworm.com/read/304730/13788049
cpp main.cpp
#include "Gold.cpp"
#include
#include
void main()
{
int i,j,p,m,mSeqLength,sequence[100000];
int length;
int num1,num2,num3;
int q;
//complex comple
www.eeworm.com/read/304730/13788067
cpp main.cpp
#include "Chu.cpp"
#include
#include
void main()
{
int i,j,p,m,mSeqLength,sequence[1000];
int length;
int num1,num2,num3;
//complex complex1;
FI