How to assign a structure member to a variable in C
Hi everyone i'm new in C and i want to know if it's possible to assign a structure member to a variable in C like below
struct User
{
int ID;
char name[10];
}users[10];
int uid,
char userName;
uid = users[i].ID;
userName = user[i].name;
trying to do something like above i got a warning that "assignment makes integer from pointer without a cast"
and when compile i run since it was a warning not an error it run but then i can't print the value of uid or userName the system will just quit;
Please i know many will ask while can't i just access the structure member directly and print it i know i can do it that way but what i actually want to to do is to see a way i can assign the value to variable
c struct
asked Dec 30 '18 at 14:43
samsam
3711621
|
show 1 more comment
Hi everyone i'm new in C and i want to know if it's possible to assign a structure member to a variable in C like below
struct User
{
int ID;
char name[10];
}users[10];
int uid,
char userName;
uid = users[i].ID;
userName = user[i].name;
trying to do something like above i got a warning that "assignment makes integer from pointer without a cast"
and when compile i run since it was a warning not an error it run but then i can't print the value of uid or userName the system will just quit;
Please i know many will ask while can't i just access the structure member directly and print it i know i can do it that way but what i actually want to to do is to see a way i can assign the value to variable
c struct
asked Dec 30 '18 at 14:43
samsam
3711621
Without seeing a Minimal, Complete, and Verifiable example I can only guess thatuser[i].nameis a pointer (i.e.char *)? Acharis not the same as achar *.
– Some programmer dude
Dec 30 '18 at 14:46
well the question has very less details. but change thechar userNametochar *userNameif the user[i].name is a string.
– Revolver
Dec 30 '18 at 14:51
@ Revolver thanks for helping i will edit my question and post the structure
– sam
Dec 30 '18 at 14:56
A name cannot be a single character. Membernameis either achar *or achar. In both cases you cannot assign to a variable of typechar.
– Gerhardh
Dec 30 '18 at 14:58
Thanks to everyone your suggestion work right i define the variable name as a pointer and that solve the problem
– sam
Dec 30 '18 at 15:03
|
show 1 more comment
Hi everyone i'm new in C and i want to know if it's possible to assign a structure member to a variable in C like below
struct User
{
int ID;
char name[10];
}users[10];
int uid,
char userName;
uid = users[i].ID;
userName = user[i].name;
trying to do something like above i got a warning that "assignment makes integer from pointer without a cast"
and when compile i run since it was a warning not an error it run but then i can't print the value of uid or userName the system will just quit;
Please i know many will ask while can't i just access the structure member directly and print it i know i can do it that way but what i actually want to to do is to see a way i can assign the value to variable
c struct
asked Dec 30 '18 at 14:43
samsam
3711621
Hi everyone i'm new in C and i want to know if it's possible to assign a structure member to a variable in C like below
struct User
{
int ID;
char name[10];
}users[10];
int uid,
char userName;
uid = users[i].ID;
userName = user[i].name;
trying to do something like above i got a warning that "assignment makes integer from pointer without a cast"
and when compile i run since it was a warning not an error it run but then i can't print the value of uid or userName the system will just quit;
Please i know many will ask while can't i just access the structure member directly and print it i know i can do it that way but what i actually want to to do is to see a way i can assign the value to variable
c struct
c struct
asked Dec 30 '18 at 14:43
samsam
3711621
asked Dec 30 '18 at 14:43
samsam
3711621
edited Dec 30 '18 at 14:58
sam
asked Dec 30 '18 at 14:43
samsam
3711621
asked Dec 30 '18 at 14:43
samsam
3711621
asked Dec 30 '18 at 14:43
samsam
3711621
3711621
Without seeing a Minimal, Complete, and Verifiable example I can only guess thatuser[i].nameis a pointer (i.e.char *)? Acharis not the same as achar *.
– Some programmer dude
Dec 30 '18 at 14:46
well the question has very less details. but change thechar userNametochar *userNameif the user[i].name is a string.
– Revolver
Dec 30 '18 at 14:51
@ Revolver thanks for helping i will edit my question and post the structure
– sam
Dec 30 '18 at 14:56
A name cannot be a single character. Membernameis either achar *or achar. In both cases you cannot assign to a variable of typechar.
– Gerhardh
Dec 30 '18 at 14:58
Thanks to everyone your suggestion work right i define the variable name as a pointer and that solve the problem
– sam
Dec 30 '18 at 15:03
|
show 1 more comment
Without seeing a Minimal, Complete, and Verifiable example I can only guess thatuser[i].nameis a pointer (i.e.char *)? Acharis not the same as achar *.
– Some programmer dude
Dec 30 '18 at 14:46
well the question has very less details. but change thechar userNametochar *userNameif the user[i].name is a string.
– Revolver
Dec 30 '18 at 14:51
@ Revolver thanks for helping i will edit my question and post the structure
– sam
Dec 30 '18 at 14:56
A name cannot be a single character. Membernameis either achar *or achar. In both cases you cannot assign to a variable of typechar.
– Gerhardh
Dec 30 '18 at 14:58
Thanks to everyone your suggestion work right i define the variable name as a pointer and that solve the problem
– sam
Dec 30 '18 at 15:03
Without seeing a Minimal, Complete, and Verifiable example I can only guess that
user[i].name is a pointer (i.e. char * )? A char is not the same as a char *.– Some programmer dude
Dec 30 '18 at 14:46
Without seeing a Minimal, Complete, and Verifiable example I can only guess that
user[i].name is a pointer (i.e. char * )? A char is not the same as a char *.– Some programmer dude
Dec 30 '18 at 14:46
well the question has very less details. but change the
char userName to char *userName if the user[i].name is a string.– Revolver
Dec 30 '18 at 14:51
well the question has very less details. but change the
char userName to char *userName if the user[i].name is a string.– Revolver
Dec 30 '18 at 14:51
@ Revolver thanks for helping i will edit my question and post the structure
– sam
Dec 30 '18 at 14:56
@ Revolver thanks for helping i will edit my question and post the structure
– sam
Dec 30 '18 at 14:56
A name cannot be a single character. Member
name is either a char * or a char. In both cases you cannot assign to a variable of type char.– Gerhardh
Dec 30 '18 at 14:58
A name cannot be a single character. Member
name is either a char * or a char. In both cases you cannot assign to a variable of type char.– Gerhardh
Dec 30 '18 at 14:58
Thanks to everyone your suggestion work right i define the variable name as a pointer and that solve the problem
– sam
Dec 30 '18 at 15:03
Thanks to everyone your suggestion work right i define the variable name as a pointer and that solve the problem
– sam
Dec 30 '18 at 15:03
|
show 1 more comment
1 Answer
1
active
oldest
votes
The problematic statement is
userName = user[i].name;
| |
char type char array type
firstly userName is declared as character variable and a character variable can't holds char array i.e user[i].name.
Change this
char userName;
to
char userName[10];
and then copy using strcpy() or strncpy(). For e.g
strcpy(userName, user[i].name);
answered Dec 30 '18 at 15:08
AchalAchal
9,1882830
2
Never use strncpy (unless you have a very good reason that you can explain when asked about it). strncpy is a time bomb waiting to blow up in your face at any time.
– gnasher729
Dec 30 '18 at 15:21
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The problematic statement is
userName = user[i].name;
| |
char type char array type
firstly userName is declared as character variable and a character variable can't holds char array i.e user[i].name.
Change this
char userName;
to
char userName[10];
and then copy using strcpy() or strncpy(). For e.g
strcpy(userName, user[i].name);
answered Dec 30 '18 at 15:08
AchalAchal
9,1882830
2
Never use strncpy (unless you have a very good reason that you can explain when asked about it). strncpy is a time bomb waiting to blow up in your face at any time.
– gnasher729
Dec 30 '18 at 15:21
add a comment |
The problematic statement is
userName = user[i].name;
| |
char type char array type
firstly userName is declared as character variable and a character variable can't holds char array i.e user[i].name.
Change this
char userName;
to
char userName[10];
and then copy using strcpy() or strncpy(). For e.g
strcpy(userName, user[i].name);
answered Dec 30 '18 at 15:08
AchalAchal
9,1882830
2
Never use strncpy (unless you have a very good reason that you can explain when asked about it). strncpy is a time bomb waiting to blow up in your face at any time.
– gnasher729
Dec 30 '18 at 15:21
add a comment |
The problematic statement is
userName = user[i].name;
| |
char type char array type
firstly userName is declared as character variable and a character variable can't holds char array i.e user[i].name.
Change this
char userName;
to
char userName[10];
and then copy using strcpy() or strncpy(). For e.g
strcpy(userName, user[i].name);
answered Dec 30 '18 at 15:08
AchalAchal
9,1882830
The problematic statement is
userName = user[i].name;
| |
char type char array type
firstly userName is declared as character variable and a character variable can't holds char array i.e user[i].name.
Change this
char userName;
to
char userName[10];
and then copy using strcpy() or strncpy(). For e.g
strcpy(userName, user[i].name);
answered Dec 30 '18 at 15:08
AchalAchal
9,1882830
edited Dec 30 '18 at 17:57
answered Dec 30 '18 at 15:08
AchalAchal
9,1882830
answered Dec 30 '18 at 15:08
AchalAchal
9,1882830
answered Dec 30 '18 at 15:08
AchalAchal
9,1882830
9,1882830
2
Never use strncpy (unless you have a very good reason that you can explain when asked about it). strncpy is a time bomb waiting to blow up in your face at any time.
– gnasher729
Dec 30 '18 at 15:21
add a comment |
2
Never use strncpy (unless you have a very good reason that you can explain when asked about it). strncpy is a time bomb waiting to blow up in your face at any time.
– gnasher729
Dec 30 '18 at 15:21
2
2
Never use strncpy (unless you have a very good reason that you can explain when asked about it). strncpy is a time bomb waiting to blow up in your face at any time.
– gnasher729
Dec 30 '18 at 15:21
Never use strncpy (unless you have a very good reason that you can explain when asked about it). strncpy is a time bomb waiting to blow up in your face at any time.
– gnasher729
Dec 30 '18 at 15:21
add a comment |
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Without seeing a Minimal, Complete, and Verifiable example I can only guess that
user[i].nameis a pointer (i.e.char *)? Acharis not the same as achar *.– Some programmer dude
Dec 30 '18 at 14:46
well the question has very less details. but change the
char userNametochar *userNameif the user[i].name is a string.– Revolver
Dec 30 '18 at 14:51
@ Revolver thanks for helping i will edit my question and post the structure
– sam
Dec 30 '18 at 14:56
A name cannot be a single character. Member
nameis either achar *or achar. In both cases you cannot assign to a variable of typechar.– Gerhardh
Dec 30 '18 at 14:58
Thanks to everyone your suggestion work right i define the variable name as a pointer and that solve the problem
– sam
Dec 30 '18 at 15:03