I usually use printf("%-8d",a); for example for 8 spaces after (and including) an integer.
My code:
#include <stdio.h>
#include <string.h>
int main()
{
int a = 10;
char b = "Hello";
}
How can I print: '#10-Hello '
with 16 spaces (8 is the integer and the string, and 8 spaces after)?
Do it in two steps. First combine the number and string with sprintf(), then print that resulting string in a 16-character field.
int a = 10;
char *b = "Hello";
char temp[20];
sprintf(temp, "#%d-%s", a, b);
printf("%-16s", temp);
snprintf() to ward off buffer overruns with updated code?A tab is 8 spaces, so, you can add \t\t The below is a super basic way to print what you wanted.
printf('#' + a + '-' + b + '\t\t');
I'm not as familiar with the syntax of C so it may be :
printf('#', a, '-', b, '\t\t');
Also, as mentioned in a previous answer, "Hello" is not a char but either an array of char or a String.
#include <stdio.h>
#include <string.h>
int main()
{
int a = 10;
char b[] = "Hello";
printf("#%d-%-17s",a,b);
}
this should get the job done, adjust your spacing as needed
Could do this with 2 printf()s. Use the return value of the first to know its print length, then print spaces needed to form a width of 16. No temporary buffer needed.
#include <assert.h>
#include <stdio.h>
int main(void) {
int width = 16;
int a = 10;
char *b = "Hello"; // Use char *
int len = printf("#%d-%s", a, b);
assert(len <= width && len >= 0);
printf("%*s", width - len, ""); // Print spaces
}
char b = "Hello"is a syntactic error, use array of char instead' ') but any characters in the output.#a-bin a 16-character field with space padding.