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any number of parameters. No limit!
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How many arguments pass in function?
zero or more it can be fixed or variable (printf is an example)
Creating a program using switch statement and if and else statement?
Not that difficult... int main (int argc, char **argv) { if (argc==1) printf ("No arguments\n"); else printf ("%d arguments\n", argc-1); goto END; END: return 0; }
Wap for swapping values of two variables using pointers as arguments to functions?
# include<stdio.h> # include<conio.h> void main() { clrscr(); int a,b; printf ("Enter the first value:"); scanf ("%d",& a ); printf ("Enter the second value:"); scanf ("%d",& b ); printf ("\n\nBefor swaping the values "); printf ("\nThe first value is %d",a); printf ("\nThe second value is %d",b); printf ("\n\nAfter swaping the values "); printf ("\nThe first value is %d",b); printf ("\nThe second value is %d",a); }
Is printf keyword?
== == What is printf in c or prototype of printf with example (q) What is prototype of printf function ? Explain each term. Ans: Prototype of printf function is : int printf( const char *format ,…) Explanation of each term : Parameter of printf function is : (three continuous dots) : It is called ellipsis. It indicates the variable number of arguments. Example of ellipsis: #include void ellipsis(int a,...); void main() { int a=5,b=10; clrscr(); ellipsis(a,b); getch(); } void ellipsis(int a,...) { printf("%d ",a); } Output:5 So printf function can have any number of variables as an argument.
C programming diamond shape for loops Problem and output sample is in the picture?
#include<stdio.h> main() { int i; for(i=1;i<=1;i++) { printf("*",i); } printf("\n"); for(i=1;i<=3;i++) { printf("*",i); } printf("\n"); for(i=1;i<=5;i++) { printf("*",i); } printf("\n"); for(i=1;i<=3;i++) { printf("*",i); } printf("\n"); for(i=1;i<=1;i++) { printf("*",i); } }
How many arguments pass in function?
zero or more it can be fixed or variable (printf is an example)
Creating a program using switch statement and if and else statement?
Not that difficult... int main (int argc, char **argv) { if (argc==1) printf ("No arguments\n"); else printf ("%d arguments\n", argc-1); goto END; END: return 0; }
What is the difference between echo and printf command of UNIX?
Echo merely repeats its command line arguments as stated. The 'printf' command allows for formatted print and a more exacting method of printing out information. Printf requires you to specify the format of what you want to print; echo does not have this ability.
Wap for swapping values of two variables using pointers as arguments to functions?
# include<stdio.h> # include<conio.h> void main() { clrscr(); int a,b; printf ("Enter the first value:"); scanf ("%d",& a ); printf ("Enter the second value:"); scanf ("%d",& b ); printf ("\n\nBefor swaping the values "); printf ("\nThe first value is %d",a); printf ("\nThe second value is %d",b); printf ("\n\nAfter swaping the values "); printf ("\nThe first value is %d",b); printf ("\nThe second value is %d",a); }
Can you use nested printf?
printf ("nested printf returned %d\n", printf ("inner printf\n"));
How do you get your boyfriend back if you have finished over too many arguments?
Too many arguments now = Too many arguments later
Is printf keyword?
== == What is printf in c or prototype of printf with example (q) What is prototype of printf function ? Explain each term. Ans: Prototype of printf function is : int printf( const char *format ,…) Explanation of each term : Parameter of printf function is : (three continuous dots) : It is called ellipsis. It indicates the variable number of arguments. Example of ellipsis: #include void ellipsis(int a,...); void main() { int a=5,b=10; clrscr(); ellipsis(a,b); getch(); } void ellipsis(int a,...) { printf("%d ",a); } Output:5 So printf function can have any number of variables as an argument.
C programming diamond shape for loops Problem and output sample is in the picture?
#include<stdio.h> main() { int i; for(i=1;i<=1;i++) { printf("*",i); } printf("\n"); for(i=1;i<=3;i++) { printf("*",i); } printf("\n"); for(i=1;i<=5;i++) { printf("*",i); } printf("\n"); for(i=1;i<=3;i++) { printf("*",i); } printf("\n"); for(i=1;i<=1;i++) { printf("*",i); } }
What is an arity?
An arity is a number of arguments or operands a function or operation takes.
What Is Arity?
An arity is a number of arguments or operands a function or operation takes.
What is an adicity?
An adicity is the number of arguments or operands a function or operation takes.
Can you write a function similar to scanf?
//program for myprintf using variable arguments #include <stdio.h> #include <stdlib.h> #include <unistd.h> #include <stdarg.h> int compare(const void * var1,const void * var2) { if(*(int *)var1 > *(int *)var2) { return(0); } else{ return(1); } } int myPrintf(const char * output ,...) { va_list arguments; int stringlength,stringIndx; char *s; int d; char c; stringlength = strlen(output); va_start(arguments,output); for(stringIndx = 0;stringIndx < stringlength;stringIndx++) { if(output[stringIndx] == '%') { stringIndx++; switch(output[stringIndx]) { case 's': /* string */ s = va_arg(arguments, char *); printf("%s\n", s); break; case 'd': /* int */ d = va_arg(arguments, int); printf("%d\n", d); break; case 'c': /* char */ default: /* need a cast here since va_arg only takes fully promoted types */ c = (char) va_arg(arguments, char); printf("%c\n", c); break; } } else{ printf("%c",output[stringIndx]); } } } void main() { unsigned int au32Nos[10] = {32,44.,55,66,11,8,9,7,9,10}; qsort(au32Nos,10,4,compare); myPrintf(" %d %d %d %d %d %d %d %d %d %d",au32Nos[0],au32Nos[1],au32Nos[2],au32Nos[3],au32Nos[4],au32Nos[5],au32Nos[6],au32Nos[7],au32Nos[8],au32Nos[9]); }
