Cleaning a static char array after use

I want my program to clean the static chars array for a new input I've tried with memset but it doesn't work or i'm doing it wrong. please some advice. Thanks in advance.

const char * password() {

     static int i = 0;

     static char pwd[STRING_LEN], c = '';

     static char return_buffer[255];



     memset(pwd, 0, sizeof pwd);     //<---

     memset(return_buffer, 0, sizeof return_buffer);     //<--



     printf("Password: ");

     while (i < STRING_LEN){

         pwd[i] = getch();

         c = pwd[i];

         if(c == 13) { break; }

         i++;

     }

     pwd[i] = '';



     snprintf(return_buffer, sizeof(return_buffer), "%s", pwd);

     printf("n");

     return return_buffer;

}

asked Nov 12 '18 at 18:01

Gabriel Vieira

2216

1

In what way is it not working?

– Christian Gibbons
Nov 12 '18 at 18:02

You'll write the '' outside the array if the user enters STRING_LEN characters. You need while (i < STRING_LEN - 1) to leave room for the null.

– Barmar
Nov 12 '18 at 18:05

1

if(c == 13) looks suspicious.

– Christian Gibbons
Nov 12 '18 at 18:07

That depends on the platform.

– Swordfish
Nov 12 '18 at 18:09

1

13 is not 'n' but 'r' (carriage return). Since you use getch() i suspect you are on windows where a "newline" consists of 'r' and 'n', so checking for 'r' (= 13) is the right(tm) thing to do. Neither, 'r' nor 'n' should result in a crash, though.

– Swordfish
Nov 12 '18 at 19:08

|
show 1 more comment

I want my program to clean the static chars array for a new input I've tried with memset but it doesn't work or i'm doing it wrong. please some advice. Thanks in advance.

const char * password() {

     static int i = 0;

     static char pwd[STRING_LEN], c = '';

     static char return_buffer[255];



     memset(pwd, 0, sizeof pwd);     //<---

     memset(return_buffer, 0, sizeof return_buffer);     //<--



     printf("Password: ");

     while (i < STRING_LEN){

         pwd[i] = getch();

         c = pwd[i];

         if(c == 13) { break; }

         i++;

     }

     pwd[i] = '';



     snprintf(return_buffer, sizeof(return_buffer), "%s", pwd);

     printf("n");

     return return_buffer;

}

asked Nov 12 '18 at 18:01

Gabriel Vieira

2216

1

In what way is it not working?

– Christian Gibbons
Nov 12 '18 at 18:02

You'll write the '' outside the array if the user enters STRING_LEN characters. You need while (i < STRING_LEN - 1) to leave room for the null.

– Barmar
Nov 12 '18 at 18:05

1

if(c == 13) looks suspicious.

– Christian Gibbons
Nov 12 '18 at 18:07

That depends on the platform.

– Swordfish
Nov 12 '18 at 18:09

1

13 is not 'n' but 'r' (carriage return). Since you use getch() i suspect you are on windows where a "newline" consists of 'r' and 'n', so checking for 'r' (= 13) is the right(tm) thing to do. Neither, 'r' nor 'n' should result in a crash, though.

– Swordfish
Nov 12 '18 at 19:08

|
show 1 more comment

I want my program to clean the static chars array for a new input I've tried with memset but it doesn't work or i'm doing it wrong. please some advice. Thanks in advance.

const char * password() {

     static int i = 0;

     static char pwd[STRING_LEN], c = '';

     static char return_buffer[255];



     memset(pwd, 0, sizeof pwd);     //<---

     memset(return_buffer, 0, sizeof return_buffer);     //<--



     printf("Password: ");

     while (i < STRING_LEN){

         pwd[i] = getch();

         c = pwd[i];

         if(c == 13) { break; }

         i++;

     }

     pwd[i] = '';



     snprintf(return_buffer, sizeof(return_buffer), "%s", pwd);

     printf("n");

     return return_buffer;

}

asked Nov 12 '18 at 18:01

Gabriel Vieira

2216

I want my program to clean the static chars array for a new input I've tried with memset but it doesn't work or i'm doing it wrong. please some advice. Thanks in advance.

const char * password() {

     static int i = 0;

     static char pwd[STRING_LEN], c = '';

     static char return_buffer[255];



     memset(pwd, 0, sizeof pwd);     //<---

     memset(return_buffer, 0, sizeof return_buffer);     //<--



     printf("Password: ");

     while (i < STRING_LEN){

         pwd[i] = getch();

         c = pwd[i];

         if(c == 13) { break; }

         i++;

     }

     pwd[i] = '';



     snprintf(return_buffer, sizeof(return_buffer), "%s", pwd);

     printf("n");

     return return_buffer;

}

c arrays clear memset

asked Nov 12 '18 at 18:01

Gabriel Vieira

2216

asked Nov 12 '18 at 18:01

Gabriel Vieira

2216

asked Nov 12 '18 at 18:01

Gabriel Vieira

2216

asked Nov 12 '18 at 18:01

Gabriel Vieira

2216

asked Nov 12 '18 at 18:01

Gabriel Vieira

2216

1

In what way is it not working?

– Christian Gibbons
Nov 12 '18 at 18:02

You'll write the '' outside the array if the user enters STRING_LEN characters. You need while (i < STRING_LEN - 1) to leave room for the null.

– Barmar
Nov 12 '18 at 18:05

1

if(c == 13) looks suspicious.

– Christian Gibbons
Nov 12 '18 at 18:07

That depends on the platform.

– Swordfish
Nov 12 '18 at 18:09

1

13 is not 'n' but 'r' (carriage return). Since you use getch() i suspect you are on windows where a "newline" consists of 'r' and 'n', so checking for 'r' (= 13) is the right(tm) thing to do. Neither, 'r' nor 'n' should result in a crash, though.

– Swordfish
Nov 12 '18 at 19:08

|
show 1 more comment

1

In what way is it not working?

– Christian Gibbons
Nov 12 '18 at 18:02

You'll write the '' outside the array if the user enters STRING_LEN characters. You need while (i < STRING_LEN - 1) to leave room for the null.

– Barmar
Nov 12 '18 at 18:05

1

if(c == 13) looks suspicious.

– Christian Gibbons
Nov 12 '18 at 18:07

That depends on the platform.

– Swordfish
Nov 12 '18 at 18:09

1

13 is not 'n' but 'r' (carriage return). Since you use getch() i suspect you are on windows where a "newline" consists of 'r' and 'n', so checking for 'r' (= 13) is the right(tm) thing to do. Neither, 'r' nor 'n' should result in a crash, though.

– Swordfish
Nov 12 '18 at 19:08

In what way is it not working?

– Christian Gibbons
Nov 12 '18 at 18:02

You'll write the '' outside the array if the user enters STRING_LEN characters. You need while (i < STRING_LEN - 1) to leave room for the null.

– Barmar
Nov 12 '18 at 18:05

if(c == 13) looks suspicious.

– Christian Gibbons
Nov 12 '18 at 18:07

That depends on the platform.

– Swordfish
Nov 12 '18 at 18:09

13 is not 'n' but 'r' (carriage return). Since you use getch() i suspect you are on windows where a "newline" consists of 'r' and 'n', so checking for 'r' (= 13) is the right(tm) thing to do. Neither, 'r' nor 'n' should result in a crash, though.

– Swordfish
Nov 12 '18 at 19:08

|
show 1 more comment

1 Answer
1

active

oldest

votes

Besides that there is no need to "clean" the arrays since you are overwriting their content with the result of getch() and a '' ... you never reset i.

i is a static int that only will get initialized the first time, execution passes its point of definition. Also, there is no need for i and pwd to be static variables.

edited Nov 12 '18 at 18:13

answered Nov 12 '18 at 18:05

Swordfish

9,06811335

Why does he need to reset i?

– Barmar
Nov 12 '18 at 18:07

Because he surely doesn't want input being written behind former input at a 2nd, 3rd, ... call to the function.

– Swordfish
Nov 12 '18 at 18:08

3

i is a static int. Subsequent calls to the function will keep incrementing without resetting it back to 0.

– Christian Gibbons
Nov 12 '18 at 18:08

Yeah, just noticed the static declaration.

– Barmar
Nov 12 '18 at 18:08

The answer should go into more detail about how static variable initialization works. There's also no need for i and pwd to be static at all. The only variable that needs to be static is return_buffer, since it's being returned.

– Barmar
Nov 12 '18 at 18:10

|
show 1 more comment

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1 Answer
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active

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active

oldest

votes

Besides that there is no need to "clean" the arrays since you are overwriting their content with the result of getch() and a '' ... you never reset i.

i is a static int that only will get initialized the first time, execution passes its point of definition. Also, there is no need for i and pwd to be static variables.

edited Nov 12 '18 at 18:13

answered Nov 12 '18 at 18:05

Swordfish

9,06811335

Why does he need to reset i?

– Barmar
Nov 12 '18 at 18:07

Because he surely doesn't want input being written behind former input at a 2nd, 3rd, ... call to the function.

– Swordfish
Nov 12 '18 at 18:08

3

i is a static int. Subsequent calls to the function will keep incrementing without resetting it back to 0.

– Christian Gibbons
Nov 12 '18 at 18:08

Yeah, just noticed the static declaration.

– Barmar
Nov 12 '18 at 18:08

The answer should go into more detail about how static variable initialization works. There's also no need for i and pwd to be static at all. The only variable that needs to be static is return_buffer, since it's being returned.

– Barmar
Nov 12 '18 at 18:10

|
show 1 more comment

Besides that there is no need to "clean" the arrays since you are overwriting their content with the result of getch() and a '' ... you never reset i.

i is a static int that only will get initialized the first time, execution passes its point of definition. Also, there is no need for i and pwd to be static variables.

edited Nov 12 '18 at 18:13

answered Nov 12 '18 at 18:05

Swordfish

9,06811335

Why does he need to reset i?

– Barmar
Nov 12 '18 at 18:07

Because he surely doesn't want input being written behind former input at a 2nd, 3rd, ... call to the function.

– Swordfish
Nov 12 '18 at 18:08

3

i is a static int. Subsequent calls to the function will keep incrementing without resetting it back to 0.

– Christian Gibbons
Nov 12 '18 at 18:08

Yeah, just noticed the static declaration.

– Barmar
Nov 12 '18 at 18:08

The answer should go into more detail about how static variable initialization works. There's also no need for i and pwd to be static at all. The only variable that needs to be static is return_buffer, since it's being returned.

– Barmar
Nov 12 '18 at 18:10

|
show 1 more comment

Besides that there is no need to "clean" the arrays since you are overwriting their content with the result of getch() and a '' ... you never reset i.

i is a static int that only will get initialized the first time, execution passes its point of definition. Also, there is no need for i and pwd to be static variables.

edited Nov 12 '18 at 18:13

answered Nov 12 '18 at 18:05

Swordfish

9,06811335

Besides that there is no need to "clean" the arrays since you are overwriting their content with the result of getch() and a '' ... you never reset i.

i is a static int that only will get initialized the first time, execution passes its point of definition. Also, there is no need for i and pwd to be static variables.

edited Nov 12 '18 at 18:13

answered Nov 12 '18 at 18:05

Swordfish

9,06811335

edited Nov 12 '18 at 18:13

answered Nov 12 '18 at 18:05

Swordfish

9,06811335

answered Nov 12 '18 at 18:05

Swordfish

9,06811335

answered Nov 12 '18 at 18:05

Swordfish

9,06811335

Why does he need to reset i?

– Barmar
Nov 12 '18 at 18:07

Because he surely doesn't want input being written behind former input at a 2nd, 3rd, ... call to the function.

– Swordfish
Nov 12 '18 at 18:08

3

i is a static int. Subsequent calls to the function will keep incrementing without resetting it back to 0.

– Christian Gibbons
Nov 12 '18 at 18:08

Yeah, just noticed the static declaration.

– Barmar
Nov 12 '18 at 18:08

The answer should go into more detail about how static variable initialization works. There's also no need for i and pwd to be static at all. The only variable that needs to be static is return_buffer, since it's being returned.

– Barmar
Nov 12 '18 at 18:10

|
show 1 more comment

Why does he need to reset i?

– Barmar
Nov 12 '18 at 18:07

Because he surely doesn't want input being written behind former input at a 2nd, 3rd, ... call to the function.

– Swordfish
Nov 12 '18 at 18:08

3

i is a static int. Subsequent calls to the function will keep incrementing without resetting it back to 0.

– Christian Gibbons
Nov 12 '18 at 18:08

Yeah, just noticed the static declaration.

– Barmar
Nov 12 '18 at 18:08

The answer should go into more detail about how static variable initialization works. There's also no need for i and pwd to be static at all. The only variable that needs to be static is return_buffer, since it's being returned.

– Barmar
Nov 12 '18 at 18:10

Why does he need to reset i?

– Barmar
Nov 12 '18 at 18:07

Because he surely doesn't want input being written behind former input at a 2nd, 3rd, ... call to the function.

– Swordfish
Nov 12 '18 at 18:08

i is a static int. Subsequent calls to the function will keep incrementing without resetting it back to 0.

– Christian Gibbons
Nov 12 '18 at 18:08

Yeah, just noticed the static declaration.

– Barmar
Nov 12 '18 at 18:08

The answer should go into more detail about how static variable initialization works. There's also no need for i and pwd to be static at all. The only variable that needs to be static is return_buffer, since it's being returned.

– Barmar
Nov 12 '18 at 18:10

|
show 1 more comment

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