How do I append blank spaces to the end of a string using printf?
I can't find any examples where spaces are appended to the right. A similar question I found use printf to add spaces to the left of the string instead.
Use negative numbers to left-align (i.e. "pad" to the right).
#include <stdio.h>
int main() {
const char *s = "hello";
printf("%30sworld\n", s);
printf("%-30sworld\n", s);
}
This prints
helloworld
hello world
Related
Rather simple question about the fgets()-function.
As I understood, the fgets() function wants to add newline: \n and NULL: \0 at the end.
So. If i give input: car
Then: char line[5] = {c,a,r,\n,\0} , isn't that so?
But, the question is: what if I give carB as input?
Is it then: char line[5] = {c,a,r,b,\0} ?
So, does fgets() always force the NULL character inside the char line?
int main(void){
char line[5];
printf("Give an input. Please give max length of 3 characters so fgets can put newline and NULL inside ");
fgets(line, 5, stdin);
return 0;
}
After printing, if I write the name I get:
Insert a name :Andrea
There is no space before Andrea, even if I put the space in the output. How can I add a space before writing the name?
#include <stdio.h>
int main() {
int string[8];
printf("\nInsert a name : ");
printf(" ");
scanf("%s", string);
printf("\nThe name is : %s", string);
return 0;
}
The code that you provided does not compile because the variable string is an array of integers (and not a string as you might think). And when you input the value of it, you're telling the compiler to expect a string ("%s" in scanf("%s", string")).
Change your string declaration to char string[8] and it should work (with the spacing you want):
Note that the value that goes into string will not contain more than 8 characters.
Actually, you did a small mistake in your code, you declared the string to be an array of int & from the code scanf("%s", string") the compiler expects from the user to enter a string as it contains %s as the format specifier. Hence, the code doesn't compile.
In order to correct this, you should declare string as char string[8] (string containing 8 characters).
The correct code is given below :
#include <stdio.h>
int main() {
char string[8];
printf("\nInsert a name : ");
scanf("%s", string);
printf("\nThe name is : %s", string);
return 0;
}
From the above correction your problem for space between input and output will be solved!
Here, is the output which is received from this correction :
I hope so this explanation will be helpful for you!
If any problem persists then feel free to ask in comments! ;-)
I'm writing a hand-written lexer for a small language but have one weird requirement that I'm not sure how to handle.
I need to be able to support the notion of delimited strings where the delimiter could be any char. eg. strings are most likely to be delimited using double quotes (eg. "hello") but it could just as easily be /hello/ or ,hello,
eg. some sample input lines might be:
x = /abc/
y = "abc" + ,def,
z = zabcz
The last case is a bit pathological, but technically possible.
I'm trying work out if there's any way I can do this in the tokenization phase in the general case? Any thoughts or suggestions would be grand.
Here are solutions in c++ and js.
c++
#include "vector"
#include "string"
#include "iostream"
using namespace std;
// Lexically Analyze method
auto lex_argument(string code){
// Define variables
size_t equal_location;
int counter = 0;
auto variable;
string variable_name;
auto variable_info[2]
string code_for_inspection;
/* In the case of a variable , these two characters will hold the beginning and end of the string */
char string_variable_characters[2];
equal_location = code.find("=",0,code.length());
variable_value = code.substr(equal_location + 2,code.length());
variable_name = code.substr(code.begin(),equal_location - 2);
variable_info[0] = variable_name;
string_variable_characters[0] = (char) variable_value.substr(0,1);
string_variable_characters[1] = (char)
variable_value.substr(variable_value.length() - 1,variable_value.length());
if(string_variable_charecters[0] = string_variable_charecters[1]){
variable_name.erase(0,1);
variable_value.erase(variable_value.length() - 1,variable_value.length());
variable_info[1] = variable_value;
}
return variable_info;
}
and in js:
function lex_argument(code){
var equalLocation = code.search("=");
var variableInfo = [null,null];
variableInfo[1] = code.substr(1,equalLocation - 2);
variableInfo[0] = code.substr(equalLocation,code.length);
string_delimeters = [variableInfo[0].substr(1,2),variableInfo[0].substr(variableInfo[0].length - 1,variableInfo[0].length];
return variableInfo;
}
As g-wan documented, xbuf_repl is replacing all occurrences. But my installed g-wan, running the following code, only replaced the first occurrence of the matched.
#include "gwan.h"
int main(int argc, char* argv[]){
xbuf_t *reply = get_reply(argv);
char str[ ] = "kjfdkkkkfldjfjfldkjdkkklfjworhg8kkkugpugulrghkkkr8g";
xbuf_ncat(reply, str, sizeof(str)-1);
xbuf_repl(reply, "kkk", "((()))");
return 200;
}
output is : kjfd((()))kfldjfjfldkjdkkklfjworhg8kkkugpugulrghkkkr8g
What's wrong of my code? How to work around it?
In the entity.c G-WAN example, you can see:
// escape '<' because it cuts the text
while(xbuf_replfrto(reply, pos, reply->ptr + reply->len - 13, "<", "<"));
This while() makes it obvious that one instance is replaced at a time, which is confirmed by the G-WAN documentation:
// replace the first occurence of the 'old' string by the 'new' string in the buffer
char *xbuf_repl (xbuf_t *ctx, char *old, char *new);
According to the G-WAN API documentation . . .
// replace the first occurence of the 'old' string by the 'new' string in the buffer
char *xbuf_repl (xbuf_t *ctx, char *old, char *new);
G-WAN also has this API, which (as I just learned from Gil's answer) also only replaces the first occurrence but within a range of the buffer rather than first occurrence from the beginning of the buffer . . . .
// same as above but using a range in the buffer
char *xbuf_replfrto(xbuf_t *ctx, char *beg, char *end, char *old, char *new);
Gil's answer shows how you can make use of this to replace ALL occurrences within the buffer.
Ken
Good afternoon,
I'm trying to accomplish a task that i know should be doable. however my attempts seem to fail every time. My endeavor is to learn to code in Objective -c and have been making good progress. what i would like to do is add a loop to my current application that asks at the end if i would like to run again or some thing to that regard, and reply with a yes or no. if no the program ends and if yes it jumps back to the top of the project to start all over. kinda like what i have below? forgive me please if its not quite perfect, im still getting used to programing and am finding it incredibly fun.
#include <stdio.h>
int main(void)
{
char loop = yes;
while (loop = yes)
{
.
.
.
}
printf ("would you like to continue (yes/no)/n");
scanf ("%s", loop);
}
The printf and scanf need to be moved up inside the curly braces of the while loop. Also, you want \n instead of /n in the printf. Finally, you're going to get a string back with that scanf() call, so you'll want to declare loop as a char array, and then in the while loop, check the first element of that array for a 'y' or 'n' or something like that. You might also want to look at getchar() instead of scanf() for that sort of thing.
Not compiled here, but should work:
#include <stdio.h>
int main(void)
{
char buffer[256];
do {
.
.
.
printf ("would you like to continue (yes/no)/n");
scanf ("%s", buffer);
} while (strcmp(buffer,"yes") != 0);
}
One wouldn't do anything like that in a real world application, but for demonstration purpose it should be ok.
I made your variable an array, because strings are arrays of characters in C. Length is set to 256 bytes (255 characters + 0-byte as delimiter). I changed the loop to do-while to make it run at least once. For string comparison you need to call a function. strcmp returns 0 for identical strings. Finally, the question belongs in the loop.
It is plain C though, using nothing of Objective-C.
int main() {
char A = 'n';
char B = 'y';
char Answer;
printf("Does the subject have a glazed over look? (y/n): \n");
scanf("%c",&Answer);
if (Answer=='N'||Answer=='y'|| Answer=='N'||Answer=='Y')
printf("Good\n");
else
printf("Please enter 'y' or 'n' \n ");
return 0;
}
#include <stdio.h>
int main(void)
{
avi;
char loop[10];
while (loop = yes)
{
.
.
.
}
printf ("would you like to continue (yes/no)/n");
scanf ("%s", loop);
if(strcpm(loop,"YES")==0) goto avi:
}