Using a variable instead of a constant while formatting my output and need to left align it

Question

This is a small portion of my code:

char *a;
asprintf(&a,"%%%ds",Max_FnLen);
printf(a,files[i-1]->d_name);
free(a); 
printf("%s",KNRM);
if ( (i % (180/Max_FnLen)) == 0) printf("\n");

Its running fine but I want to left align the output but it is coming right aligned by default can anyone help me with this.

To use a _variable_ to specfy the width, use e.g. `printf("%.*s", len, str)`. The `*` specifies a variable in the argument list is to be used for the width (or precission). — Paul Ogilvie, Sep 11 '17 at 11:32

gsamaras · Answer 1 · 2017-09-11T09:09:46.663

0

Use the - flag, as described in the ref:

- Left-justify within the given field width; Right justification is the default (see width sub-specifier).

edited Sep 11 '17 at 09:09

answered Sep 11 '17 at 09:04

gsamaras

It works if we are using a constant value to fix the spaces but as I am using a variable it's not working – Kyoichi Shido Sep 11 '17 at 09:26

score 0 · Accepted Answer · answered Sep 11 '17 at 09:07

0

Use "-" modifier like: printf("%-20s\n", "short line");

In your case:

char *a;
asprintf(&a,"%%-%ds",Max_FnLen);
printf(a,files[i-1]->d_name);
free(a); 
printf("%s",KNRM);
if ( (i % (180/Max_FnLen)) == 0) printf("\n");

answered Sep 11 '17 at 09:07

MrCryo

1

Thanks very much @MrCryo . It was a huge help! – Kyoichi Shido Sep 11 '17 at 09:36
`%-.*s` should be sufficient I think. (I might not remember the correct ordering, but you’ve got the idea) – 0andriy Oct 17 '21 at 15:44

2 Answers2