I get a Blinking Cursor insread of an Output when running the C program in Ubuntu.

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Here's my code for finding the distance between two points. My assignment makes it mandatory to make use of functions, hence a separate one for finding the distance. I don't get any syntax errors but only a blinking cursor when I execute the code. Please help, I'm a beginner in C.

My code:

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

float distance(float *x1, float *y1, float *x2, float *y2);

int main()
 printf("Input coordinates as x1,y1,x2,y2:n");
float x1,y1,x2,y2;
scanf("%f %f %f %f ",&x1,&y1,&x2,&y2);

printf("Coordinates are : (%f,%f) and (%f,%f)n",x1,y1,x2,y2);
printf("%f",distance (&x1,&y1,&x2,&y2));
return 0;


float distance(float *x1, float *y1, float *x2, float *y2)
 
 float d= sqrt(pow((*x1-*x2),2)+pow((*y1-*y2),2));
printf("%f", d);
return d;

while compiling:
gcc -o 1 1.c -lm
output:

Input coordinates as x1,y1,x2,y2:
1 2 3 4

That's it. The cursor keeps blinking without providing any output. Please help.

Also, I tried putting in print statements at various points. Any print after the scanf doesn't work

asked 2 hours ago

Sayantan Mukherjee

111

New contributor

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up vote
2
down vote

favorite

My code:

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

float distance(float *x1, float *y1, float *x2, float *y2);

int main()
 printf("Input coordinates as x1,y1,x2,y2:n");
float x1,y1,x2,y2;
scanf("%f %f %f %f ",&x1,&y1,&x2,&y2);

printf("Coordinates are : (%f,%f) and (%f,%f)n",x1,y1,x2,y2);
printf("%f",distance (&x1,&y1,&x2,&y2));
return 0;


float distance(float *x1, float *y1, float *x2, float *y2)
 
 float d= sqrt(pow((*x1-*x2),2)+pow((*y1-*y2),2));
printf("%f", d);
return d;

while compiling:
gcc -o 1 1.c -lm
output:

Input coordinates as x1,y1,x2,y2:
1 2 3 4

That's it. The cursor keeps blinking without providing any output. Please help.

Also, I tried putting in print statements at various points. Any print after the scanf doesn't work

asked 2 hours ago

Sayantan Mukherjee

111

New contributor

add a commentÂ |Â

up vote
2
down vote

favorite

My code:

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

float distance(float *x1, float *y1, float *x2, float *y2);

int main()
 printf("Input coordinates as x1,y1,x2,y2:n");
float x1,y1,x2,y2;
scanf("%f %f %f %f ",&x1,&y1,&x2,&y2);

printf("Coordinates are : (%f,%f) and (%f,%f)n",x1,y1,x2,y2);
printf("%f",distance (&x1,&y1,&x2,&y2));
return 0;


float distance(float *x1, float *y1, float *x2, float *y2)
 
 float d= sqrt(pow((*x1-*x2),2)+pow((*y1-*y2),2));
printf("%f", d);
return d;

while compiling:
gcc -o 1 1.c -lm
output:

Input coordinates as x1,y1,x2,y2:
1 2 3 4

That's it. The cursor keeps blinking without providing any output. Please help.

Also, I tried putting in print statements at various points. Any print after the scanf doesn't work

asked 2 hours ago

Sayantan Mukherjee

111

New contributor

My code:

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

float distance(float *x1, float *y1, float *x2, float *y2);

int main()
 printf("Input coordinates as x1,y1,x2,y2:n");
float x1,y1,x2,y2;
scanf("%f %f %f %f ",&x1,&y1,&x2,&y2);

printf("Coordinates are : (%f,%f) and (%f,%f)n",x1,y1,x2,y2);
printf("%f",distance (&x1,&y1,&x2,&y2));
return 0;


float distance(float *x1, float *y1, float *x2, float *y2)
 
 float d= sqrt(pow((*x1-*x2),2)+pow((*y1-*y2),2));
printf("%f", d);
return d;

while compiling:
gcc -o 1 1.c -lm
output:

Input coordinates as x1,y1,x2,y2:
1 2 3 4

That's it. The cursor keeps blinking without providing any output. Please help.

Also, I tried putting in print statements at various points. Any print after the scanf doesn't work

printing compiling gcc c

asked 2 hours ago

Sayantan Mukherjee

111

New contributor

asked 2 hours ago

Sayantan Mukherjee

111

New contributor

asked 2 hours ago

Sayantan Mukherjee

111

New contributor

asked 2 hours ago

Sayantan Mukherjee

111

asked 2 hours ago

Sayantan Mukherjee

111

New contributor

Sayantan Mukherjee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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2 Answers
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up vote
1
down vote

Two things:

gcc is for compiling, not running code.

-o flag specifies the name of new file to create. Because you run gcc and there's no errors, it means it compiled the C code 1.c into a file 1 that you specified with -o flag. Thus, you can just run ./1 in terminal.

Note that using numbers for executable names is bad practice and also bad naming convention. Use gcc -o distance -lm 1.c instead, and run ./distance in terminal

Note also that because we give bare name of the file, it will create executable in current working directory. That's why you need ./

answered 1 hour ago

Sergiy Kolodyazhnyy

66.5k9134294

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up vote
1
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The question you are asking is not specific to Ubuntu but to C programming. I would suggest you ask such questions at Stackoverflow -- it's for your own benefit, as well, since you'll reach a different audience.

Anyway, you should read up on the scanf () function as it is trying to match what you're typing in on the keyboard. That includes any spaces, newlines, etc. that you have in the formatting string. Then, you also need to end your input into standard in with the "Return key".

For your problem, change:

scanf("%f %f %f %f ",&x1,&y1,&x2,&y2);

to:

scanf("%f %f %f %f",&x1,&y1,&x2,&y2);

(Remove the space at the end of the formatting string.) And then it should work as expected.

Personally, I've always used scanf to read a value at a time. As shown in the example here. I admit that this was just my way of avoiding the problem you're seeing.

answered 40 mins ago

Ray

748520

Another better idea would be to use argv positional parameters instead of scanf altogether. But that's another topic entirely.
â€“Â Sergiy Kolodyazhnyy
24 mins ago

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
1
down vote

Two things:

gcc is for compiling, not running code.

-o flag specifies the name of new file to create. Because you run gcc and there's no errors, it means it compiled the C code 1.c into a file 1 that you specified with -o flag. Thus, you can just run ./1 in terminal.

Note that using numbers for executable names is bad practice and also bad naming convention. Use gcc -o distance -lm 1.c instead, and run ./distance in terminal

Note also that because we give bare name of the file, it will create executable in current working directory. That's why you need ./

answered 1 hour ago

Sergiy Kolodyazhnyy

66.5k9134294

add a commentÂ |Â

up vote
1
down vote

Two things:

gcc is for compiling, not running code.

-o flag specifies the name of new file to create. Because you run gcc and there's no errors, it means it compiled the C code 1.c into a file 1 that you specified with -o flag. Thus, you can just run ./1 in terminal.

Note that using numbers for executable names is bad practice and also bad naming convention. Use gcc -o distance -lm 1.c instead, and run ./distance in terminal

Note also that because we give bare name of the file, it will create executable in current working directory. That's why you need ./

answered 1 hour ago

Sergiy Kolodyazhnyy

66.5k9134294

add a commentÂ |Â

up vote
1
down vote

Two things:

gcc is for compiling, not running code.

-o flag specifies the name of new file to create. Because you run gcc and there's no errors, it means it compiled the C code 1.c into a file 1 that you specified with -o flag. Thus, you can just run ./1 in terminal.

Note that using numbers for executable names is bad practice and also bad naming convention. Use gcc -o distance -lm 1.c instead, and run ./distance in terminal

Note also that because we give bare name of the file, it will create executable in current working directory. That's why you need ./

answered 1 hour ago

Sergiy Kolodyazhnyy

66.5k9134294

Two things:

gcc is for compiling, not running code.

-o flag specifies the name of new file to create. Because you run gcc and there's no errors, it means it compiled the C code 1.c into a file 1 that you specified with -o flag. Thus, you can just run ./1 in terminal.

Note that using numbers for executable names is bad practice and also bad naming convention. Use gcc -o distance -lm 1.c instead, and run ./distance in terminal

Note also that because we give bare name of the file, it will create executable in current working directory. That's why you need ./

answered 1 hour ago

Sergiy Kolodyazhnyy

66.5k9134294

answered 1 hour ago

Sergiy Kolodyazhnyy

66.5k9134294

answered 1 hour ago

Sergiy Kolodyazhnyy

66.5k9134294

answered 1 hour ago

Sergiy Kolodyazhnyy

66.5k9134294

add a commentÂ |Â

up vote
1
down vote

For your problem, change:

scanf("%f %f %f %f ",&x1,&y1,&x2,&y2);

to:

scanf("%f %f %f %f",&x1,&y1,&x2,&y2);

(Remove the space at the end of the formatting string.) And then it should work as expected.

Personally, I've always used scanf to read a value at a time. As shown in the example here. I admit that this was just my way of avoiding the problem you're seeing.

answered 40 mins ago

Ray

748520

Another better idea would be to use argv positional parameters instead of scanf altogether. But that's another topic entirely.
â€“Â Sergiy Kolodyazhnyy
24 mins ago

add a commentÂ |Â

up vote
1
down vote

For your problem, change:

scanf("%f %f %f %f ",&x1,&y1,&x2,&y2);

to:

scanf("%f %f %f %f",&x1,&y1,&x2,&y2);

(Remove the space at the end of the formatting string.) And then it should work as expected.

Personally, I've always used scanf to read a value at a time. As shown in the example here. I admit that this was just my way of avoiding the problem you're seeing.

answered 40 mins ago

Ray

748520

Another better idea would be to use argv positional parameters instead of scanf altogether. But that's another topic entirely.
â€“Â Sergiy Kolodyazhnyy
24 mins ago

add a commentÂ |Â

up vote
1
down vote

For your problem, change:

scanf("%f %f %f %f ",&x1,&y1,&x2,&y2);

to:

scanf("%f %f %f %f",&x1,&y1,&x2,&y2);

(Remove the space at the end of the formatting string.) And then it should work as expected.

Personally, I've always used scanf to read a value at a time. As shown in the example here. I admit that this was just my way of avoiding the problem you're seeing.

answered 40 mins ago

Ray

748520

For your problem, change:

scanf("%f %f %f %f ",&x1,&y1,&x2,&y2);

to:

scanf("%f %f %f %f",&x1,&y1,&x2,&y2);

(Remove the space at the end of the formatting string.) And then it should work as expected.

Personally, I've always used scanf to read a value at a time. As shown in the example here. I admit that this was just my way of avoiding the problem you're seeing.

answered 40 mins ago

Ray

748520

answered 40 mins ago

Ray

748520

answered 40 mins ago

Ray

748520

answered 40 mins ago

Ray

748520

Another better idea would be to use argv positional parameters instead of scanf altogether. But that's another topic entirely.
â€“Â Sergiy Kolodyazhnyy
24 mins ago

add a commentÂ |Â

Another better idea would be to use argv positional parameters instead of scanf altogether. But that's another topic entirely.
â€“Â Sergiy Kolodyazhnyy
24 mins ago

Another better idea would be to use argv positional parameters instead of scanf altogether. But that's another topic entirely.
â€“Â Sergiy Kolodyazhnyy
24 mins ago

add a commentÂ |Â

Sayantan Mukherjee is a new contributor. Be nice, and check out our Code of Conduct.

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