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