How can i check if char pointer is points nothing after allocated
I have allocated memory for a char pointer in this code and I wanted check does it points anything:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(){
char *p = (char*) malloc(sizeof(char)*10);
if(*p == 0)
printf("The pointer points nothing !n");
}
And I checked that if it doesn't point anything. Also I print of pointer's length and it prints "3" although it prints nothing. What is reason of this and how can I check if it points nothing ?
c pointers char malloc
edited Nov 20 '18 at 9:06
Sourav Ghosh
108k14129187
asked Nov 20 '18 at 8:56
okydoky
136
add a comment |
I have allocated memory for a char pointer in this code and I wanted check does it points anything:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(){
char *p = (char*) malloc(sizeof(char)*10);
if(*p == 0)
printf("The pointer points nothing !n");
}
And I checked that if it doesn't point anything. Also I print of pointer's length and it prints "3" although it prints nothing. What is reason of this and how can I check if it points nothing ?
c pointers char malloc
edited Nov 20 '18 at 9:06
Sourav Ghosh
108k14129187
asked Nov 20 '18 at 8:56
okydoky
136
The space allocated by malloc is uninitialized. It's not possible to check whether or not something is uninitialized. You should store values there before reading them back
– M.M
Nov 20 '18 at 8:59
I don't see the code where you 'print of pointer's length and it prints "3"' - did that somehow get lost while you were posting?
– Toby Speight
Nov 20 '18 at 9:11
Sorry, i have forgotten to add this.
– okydoky
Nov 20 '18 at 9:22
add a comment |
I have allocated memory for a char pointer in this code and I wanted check does it points anything:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(){
char *p = (char*) malloc(sizeof(char)*10);
if(*p == 0)
printf("The pointer points nothing !n");
}
And I checked that if it doesn't point anything. Also I print of pointer's length and it prints "3" although it prints nothing. What is reason of this and how can I check if it points nothing ?
c pointers char malloc
edited Nov 20 '18 at 9:06
Sourav Ghosh
108k14129187
asked Nov 20 '18 at 8:56
okydoky
136
I have allocated memory for a char pointer in this code and I wanted check does it points anything:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(){
char *p = (char*) malloc(sizeof(char)*10);
if(*p == 0)
printf("The pointer points nothing !n");
}
And I checked that if it doesn't point anything. Also I print of pointer's length and it prints "3" although it prints nothing. What is reason of this and how can I check if it points nothing ?
c pointers char malloc
c pointers char malloc
edited Nov 20 '18 at 9:06
Sourav Ghosh
108k14129187
asked Nov 20 '18 at 8:56
okydoky
136
edited Nov 20 '18 at 9:06
Sourav Ghosh
108k14129187
asked Nov 20 '18 at 8:56
okydoky
136
edited Nov 20 '18 at 9:06
Sourav Ghosh
108k14129187
edited Nov 20 '18 at 9:06
Sourav Ghosh
108k14129187
edited Nov 20 '18 at 9:06
Sourav Ghosh
108k14129187
108k14129187
asked Nov 20 '18 at 8:56
okydoky
136
asked Nov 20 '18 at 8:56
okydoky
136
asked Nov 20 '18 at 8:56
okydoky
136
136
The space allocated by malloc is uninitialized. It's not possible to check whether or not something is uninitialized. You should store values there before reading them back
– M.M
Nov 20 '18 at 8:59
I don't see the code where you 'print of pointer's length and it prints "3"' - did that somehow get lost while you were posting?
– Toby Speight
Nov 20 '18 at 9:11
Sorry, i have forgotten to add this.
– okydoky
Nov 20 '18 at 9:22
add a comment |
The space allocated by malloc is uninitialized. It's not possible to check whether or not something is uninitialized. You should store values there before reading them back
– M.M
Nov 20 '18 at 8:59
I don't see the code where you 'print of pointer's length and it prints "3"' - did that somehow get lost while you were posting?
– Toby Speight
Nov 20 '18 at 9:11
Sorry, i have forgotten to add this.
– okydoky
Nov 20 '18 at 9:22
The space allocated by malloc is uninitialized. It's not possible to check whether or not something is uninitialized. You should store values there before reading them back
– M.M
Nov 20 '18 at 8:59
The space allocated by malloc is uninitialized. It's not possible to check whether or not something is uninitialized. You should store values there before reading them back
– M.M
Nov 20 '18 at 8:59
I don't see the code where you 'print of pointer's length and it prints "3"' - did that somehow get lost while you were posting?
– Toby Speight
Nov 20 '18 at 9:11
I don't see the code where you 'print of pointer's length and it prints "3"' - did that somehow get lost while you were posting?
– Toby Speight
Nov 20 '18 at 9:11
Sorry, i have forgotten to add this.
– okydoky
Nov 20 '18 at 9:22
Sorry, i have forgotten to add this.
– okydoky
Nov 20 '18 at 9:22
add a comment |
1 Answer
1
active
oldest
votes
You need to modify the check to check against the returned pointer, not the content. Something like
if( p == NULL) {
fprintf(stderr, "The pointer points nothing !n");
return 1;
}
should do. Quoting the standard regarding the return values:
The
mallocfunction returns either a null pointer or a pointer to the allocated space.
The null pointer is returned in case of a failure, otherwise the pointer returned should be non-equal to a null pointer.
That said,
The initial content of the pointed memory is indeterminate, do not attempt to verify it. Quoting from the standard,
The
mallocfunction allocates space for an object whose size is specified bysizeand
whose value is indeterminate.
That goes for your other question regarding "checking the length" of the pointer - it's pointless. Unless you have stored (write) some value into it - there's no point trying to measure the initial content, as it is indeterminate. You may eventually run into undefined behavior.
The cast for the returned pointer by
malloc()and family is superfluous. It can be totally avoided in C.sizeof(char)is guaranteed to be 1, in C. Thus, usingsizeof(char)as a multiplier, is again not needed, per se.
answered Nov 20 '18 at 8:58
Sourav Ghosh
108k14129187
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You need to modify the check to check against the returned pointer, not the content. Something like
if( p == NULL) {
fprintf(stderr, "The pointer points nothing !n");
return 1;
}
should do. Quoting the standard regarding the return values:
The
mallocfunction returns either a null pointer or a pointer to the allocated space.
The null pointer is returned in case of a failure, otherwise the pointer returned should be non-equal to a null pointer.
That said,
The initial content of the pointed memory is indeterminate, do not attempt to verify it. Quoting from the standard,
The
mallocfunction allocates space for an object whose size is specified bysizeand
whose value is indeterminate.
That goes for your other question regarding "checking the length" of the pointer - it's pointless. Unless you have stored (write) some value into it - there's no point trying to measure the initial content, as it is indeterminate. You may eventually run into undefined behavior.
The cast for the returned pointer by
malloc()and family is superfluous. It can be totally avoided in C.sizeof(char)is guaranteed to be 1, in C. Thus, usingsizeof(char)as a multiplier, is again not needed, per se.
answered Nov 20 '18 at 8:58
Sourav Ghosh
108k14129187
add a comment |
You need to modify the check to check against the returned pointer, not the content. Something like
if( p == NULL) {
fprintf(stderr, "The pointer points nothing !n");
return 1;
}
should do. Quoting the standard regarding the return values:
The
mallocfunction returns either a null pointer or a pointer to the allocated space.
The null pointer is returned in case of a failure, otherwise the pointer returned should be non-equal to a null pointer.
That said,
The initial content of the pointed memory is indeterminate, do not attempt to verify it. Quoting from the standard,
The
mallocfunction allocates space for an object whose size is specified bysizeand
whose value is indeterminate.
That goes for your other question regarding "checking the length" of the pointer - it's pointless. Unless you have stored (write) some value into it - there's no point trying to measure the initial content, as it is indeterminate. You may eventually run into undefined behavior.
The cast for the returned pointer by
malloc()and family is superfluous. It can be totally avoided in C.sizeof(char)is guaranteed to be 1, in C. Thus, usingsizeof(char)as a multiplier, is again not needed, per se.
answered Nov 20 '18 at 8:58
Sourav Ghosh
108k14129187
add a comment |
You need to modify the check to check against the returned pointer, not the content. Something like
if( p == NULL) {
fprintf(stderr, "The pointer points nothing !n");
return 1;
}
should do. Quoting the standard regarding the return values:
The
mallocfunction returns either a null pointer or a pointer to the allocated space.
The null pointer is returned in case of a failure, otherwise the pointer returned should be non-equal to a null pointer.
That said,
The initial content of the pointed memory is indeterminate, do not attempt to verify it. Quoting from the standard,
The
mallocfunction allocates space for an object whose size is specified bysizeand
whose value is indeterminate.
That goes for your other question regarding "checking the length" of the pointer - it's pointless. Unless you have stored (write) some value into it - there's no point trying to measure the initial content, as it is indeterminate. You may eventually run into undefined behavior.
The cast for the returned pointer by
malloc()and family is superfluous. It can be totally avoided in C.sizeof(char)is guaranteed to be 1, in C. Thus, usingsizeof(char)as a multiplier, is again not needed, per se.
answered Nov 20 '18 at 8:58
Sourav Ghosh
108k14129187
You need to modify the check to check against the returned pointer, not the content. Something like
if( p == NULL) {
fprintf(stderr, "The pointer points nothing !n");
return 1;
}
should do. Quoting the standard regarding the return values:
The
mallocfunction returns either a null pointer or a pointer to the allocated space.
The null pointer is returned in case of a failure, otherwise the pointer returned should be non-equal to a null pointer.
That said,
The initial content of the pointed memory is indeterminate, do not attempt to verify it. Quoting from the standard,
The
mallocfunction allocates space for an object whose size is specified bysizeand
whose value is indeterminate.
That goes for your other question regarding "checking the length" of the pointer - it's pointless. Unless you have stored (write) some value into it - there's no point trying to measure the initial content, as it is indeterminate. You may eventually run into undefined behavior.
The cast for the returned pointer by
malloc()and family is superfluous. It can be totally avoided in C.sizeof(char)is guaranteed to be 1, in C. Thus, usingsizeof(char)as a multiplier, is again not needed, per se.
answered Nov 20 '18 at 8:58
Sourav Ghosh
108k14129187
edited Nov 20 '18 at 9:06
answered Nov 20 '18 at 8:58
Sourav Ghosh
108k14129187
answered Nov 20 '18 at 8:58
Sourav Ghosh
108k14129187
answered Nov 20 '18 at 8:58
Sourav Ghosh
108k14129187
108k14129187
add a comment |
add a comment |
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The space allocated by malloc is uninitialized. It's not possible to check whether or not something is uninitialized. You should store values there before reading them back
– M.M
Nov 20 '18 at 8:59
I don't see the code where you 'print of pointer's length and it prints "3"' - did that somehow get lost while you were posting?
– Toby Speight
Nov 20 '18 at 9:11
Sorry, i have forgotten to add this.
– okydoky
Nov 20 '18 at 9:22