How to Print Multiple Variables Using Printf

Printing multiple variables in single printf() statement

For string use %s instead of %c and If your Id number is integer use %d
instead of %s

 printf("Entered string is %s with length %d and
     ID number is %s with length %d",string,i, number, j);

How can I print multiple variables using printf

Solution

printf '%s %d %s' "$infile" "$insize" "$indate" 2>&1 | tee -a "$logfile"
find "$infile" -printf ' %p %s %CY-%Cm-%Cd %CH:%CM:%.2TS \n' 2>&1 | tee -a "$logfile"

Result

ftp://ftp.ncbi.nlm.nih.gov/pub/README.ftp 2037 2015-08-12 15:47:26 ./README.ftp 2037 2019-08-24 20:32:53

man 1 printf clear and simple explanation of argument handing.

GNU Manual Table of Output Conversions for format summary

Why is the printf specifiers, %s, printing multiple variables at once?

Whats happening is the a null terminating character is implicitly being added when you initialize a char[] like below:

char myStr[] ="Ignacio";

If you want to initialize a char[] with individual characters you must initialize it like the following because the null terminating character must be explicitly added:

char myStr[] ={'I', 'g', 'n', 'c', 'i','0','\0'}).

Explanation for the Behaviour in Your Case:

Memory Layout:

[0] [1] [2] [3] [4] [5] [6] [7] [8] [9] [A] [B] [C] [D] [E] [F] [10] [11]

'D' 'i' 'e' 'g' 'o' 'D' 'i' 'e' 'g' 'o' 'I' 'g' 'n' 'a' 'c' 'i' 'o' '\0'

After you initialize your strings, the characters are arranged in memory like above. In the C language, all valid strings must have a null terminating character at there end to specificity the end of the string. The reason why is printf will just start at the address of the char* passed to it and will keep reading characters until a '\0' character is encountered. Because of this, if you pass printf an address to an array of characters that does not contain a '\0', then printf will keep marching past those characters into memory until a null terminating character is encountered. (This could result in a segmentation fault)

Explanation for Your Output: (Reference the memory layout above)

Name: DiegoIgnacio

printf will start at index [5] and will print all characters until the '\0' character, thus "DiegoIgnacio" ([5] - [11]) is the output.

Name: DiegoDiegoIgnacio

printf will start at index [0] and will print all characters until the '\0' character, thus "DiegoDiegoIgnacio" ([0] - [11]) is the output.

How to print variable multiple times with one printf statement in C

So the answer is...it is not possible to do what I wanted on the way I wanted. The best way is to use another for loop inside existing one.

Thank you all for your answers.

How to print multiple variable lines in Java

You can do it with 1 printf:

System.out.printf("First Name: %s\nLast Name: %s",firstname, lastname);

(C) Displaying multiple variables on the printf line, having some trouble

Though you names these variables int1 etc, their type is double! So either declare them as int or if you choose to use double:

Change

scanf("%d", &int1);

to

scanf("%lf", &int1);

and give them better names.

what is the correct syntax for awk's printf to insert multiple variables?

You need one format string, like %s, for every variable. So, try:

printf "%s %s\n", a, a_counter

Or:

printf "pattern=%s and count=%s\n", a, a_counter

%s converts any variable to a string. If the variable is a number, other formats, like %f or %e, give you more control over how the number is converted to a string. See man awk for details.

Examples

$ awk -v a="genie" -v a_counter=3 'BEGIN{ printf "%s %s\n", a, a_counter }' 
genie 3
$ awk -v a="genie" -v a_counter=3 'BEGIN{ printf "pattern=%s and count=%s\n", a, a_counter }' 
pattern=genie and count=3
$ awk -v a="genie" -v a_counter=3 'BEGIN{ printf "pattern=%s and count=%7.2f\n", a, a_counter }' 
pattern=genie and count=   3.00
$ awk -v a="genie" -v a_counter=3 'BEGIN{ printf "pattern=%s and count=%9.2e\n", a, a_counter }' 
pattern=genie and count= 3.00e+00

How can I print multiple character with one printf?

You can not use printf like that to print repetitive characters in Ansi C. I suggest you to use a loop like this -

#include <stdio.h>

int main()
{
    int i;
    for(i = 0; i < 10; i++) putchar('#');

    return 0;
}

Or if you have absolutely no desire to use loops, you can do something like this-

#include <stdio.h>
int main()
{
    char out[100];
    memset(out, '#', 10);
    out[10] = 0;
    printf("%s", out);

    return 0;
}

By the way, using printf like this also works-

#include <stdio.h>

int main()
{
    printf("%.*s", 10, "############################################");

    return 0;
}

printing multiple values using pointer in printf

The C standard does not define in what order the parameters get evaluated. Try using *(ptr+1) instead of *ptr++, which also can be replaced with ptr[1].

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