As I add the code of remove newline, after input very long name it will show unhandled exception. How can I remove the newline and still get the benefit of fgets at the same time?Code:#include <stdio.h> #include <stdlib.h> int main() { char name[3], *line; printf("Name: "); fgets(name, sizeof(name), stdin); // Remove newline line = strchr(name,'\n'); *line = '\0'; printf("%s\n", name); printf("sssssssssssssss\n"); system("pause"); }
Like this:
Code:if (fgets(name, sizeof(name), stdin)) { char *line = strchr(name, '\n'); if (line) { *line = '\0'; } }
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Perhaps by writing a function with the same interface?
> As I add the code of remove newline, after input very long name it will show unhandled exception. How can I remove the newline and still get the benefit of fgets at the same time?Code:char *my_fgets( char *buffer, size_t size, FILE *fp ) { char *result = fgets( buffer, size, fp ); if ( result ) { // remove \n } return result; }
If the line is very long, then it will NOT have a \n.
Without a \n, your strchr() returns NULL - and you can guess what happens when you then try *line = '\0';
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Thank you very much.
It still works fine after I remove *line = '\0';. I think *line = '\0' is redundant.Code:#include <stdio.h> #include <stdlib.h> #include <string.h> int main() { char name[3], line; printf("Name: "); fgets(name, sizeof(name), stdin); line = strchr(name,'\n'); printf("%sAAAAAAAAAAA\n", name); printf("sssssssssssssss\n"); system("pause"); }
Try typing in only one character, then press enter.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Why become like that? @@
Because now that you are not setting the newline character to '\0', it is displayed when printed, and the way to display a newline is to... print a newline.
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I wonder when face with character need to use \0 but doesn't need \0 when it is string.
What do you mean?
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How does *line = '\0' works when input 1 character such as a.
Basically, your string looks like this:
Then you replace the '\n' with '\0' to get:Code:{'a', '\n', '\0'}
The reason why Salem asked you to enter one character is that you declared name to be an array of 3 characters, so you can only enter 1 letter for the name if you want to see the effect of the replacement. Otherwise, you may end up with:Code:{'a', '\0', '\0'}
upon which no replacement happens because the '\n' is not found.Code:{'a', 'b', '\0'}
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Why only one character generates '\n'?Code:{'a', '\n', '\0'}
Because fgets stores the '\n' when there is space.
Okay, let's move away from this "one character" fixed notion. Compile and run this program:
Notice that this is the same program as you posted in post #5, except that I changed the 3 to 13, and fixed line to be a pointer instead of a char. Enter: "hello world" and observe the output. Then run the program again, entering "hello world!" and observe the output.Code:#include <stdio.h> #include <stdlib.h> #include <string.h> int main() { char name[13], *line; printf("Name: "); fgets(name, sizeof(name), stdin); line = strchr(name,'\n'); printf("%sAAAAAAAAAAA\n", name); printf("sssssssssssssss\n"); }
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Thanks a lot.
