Is there any way I can do this without using dynamic memory allocation?
#include <stdio.h>
int main() {
char* str = NULL;
scanf("%s", str);
printf("%s\n", str);
}
3 Answers
There is no straightforward way as per the C standard (C11), however, POSIX defines optional assignment-allocation character m as part of the conversion specifier, which relives the programmer from the responsibility of allocating memory.
However, under the hood, this does use dynamic memory allocation, and you need to call free() on the pointer later on.
Quoting the reference:
The
%c,%s, and%[conversion specifiers shall accept an optional assignment-allocation character'm', which shall cause a memory buffer to be allocated to hold the string converted including a terminating null character. In such a case, the argument corresponding to the conversion specifier should be a reference to a pointer variable that will receive a pointer to the allocated buffer. The system shall allocate a buffer as ifmalloc()had been called. The application shall be responsible for freeing the memory after usage. If there is insufficient memory to allocate a buffer, the function shall seterrnoto[ENOMEM]and a conversion error shall result. If the function returnsEOF, any memory successfully allocated for parameters using assignment-allocation character'm'by this call shall be freed before the function returns.
Something like
#include <stdio.h>
#include <stdlib.h>
int main() {
char* str = NULL;
scanf("%ms", &str);
printf("%s\n", str);
free(str);
return 0;
}
would do the job.
1 Comment
ryyker
GNU lib C also defines such a feature.This way:
int main()
{
char str[100];
scanf("%s", str);
printf("%s\n", str);
}
3 Comments
Tim Randall
What if the user enters a hundred or more characters?
Leo
@ryyker Well after the question title, the OP asks for a solution without using dynamic memory allocation? so...
Leo
The answer by Sourav is neat, but still uses dynamic memory allocation under the hood.
Yes, but you are going to waste memory, or have the risk of segmentanltion violation:
#include <stdio.h>
int main() {
char str[512] = {0} // as suggested by Jabberwocky
//sprintf(str, "");
//scanf("%s", str); // protecting the input from more than 511 chars
scanf("%511s", str);
printf("%s\n", str);
}
5 Comments
Tim Randall
Can you explain the risk? Is there any workaround? Perhaps another function that could be called to enter the string?
Jabberwocky
Why
sprintf(str, "");? Why not str[0] = 0; or char str[512] = {0};. And what if the user enters a string longer than 511 characters?
Lundin
So restrict the input from scanf to a fixed number of characters. Or preferably use fgets. Or better yet, forget about using stdio.h.
vfalcao
If the input is greater than the "buffer size" (say, 511 characters + ending \0) you are going to try to access memory that was not allocated to your process, and a SEGV (segmentation violation) will occur. You can limit the input, using
scanf("%511s", str);vfalcao
the idiom
sprintf(str, ""); or the suggested str[0] = 0 are unneccessary in this particular piece of code; in any case, the sprintf version seems easier to understand. Regarding the limiting, I have updated my suggested answer.
char str[100];