I'm working on my assignment for my C course, and I'm trying to take in the user's input and store it in a variable to use for later in my code. Here's what my main function looks like,
int main() {
// Variables here
char* inputLine[10];
do {
printf("Insert number....");
scanf("%s\n", inputLine);
// More stuff here
}
return 0;
}
This code gives me a bunch of warnings, warning: format specifies type 'char *' but the argument has type 'char **' [-Wformat], and if I change the variable declaration to,
char* inputLine = NULL;
When I execute my code I get a seg fault, can someone explain to me what I am doing wrong, and the differences of what happens in the memory when I'm initializing this variable?
char* inputLine[10];
--> is an array of ten pointers to char
printf's format %s expects argument of type char *, but you're providing it as type char **
Just use
char inputLine[10];
To avoid possible buffer overflow you should use
scanf("%9s", inputLine); //Notice the size with %s
9 only because C string are null terminated ('\0') so one extra byte for it goes at end
char inputLine[10];
do {
printf("Insert number....");
scanf("%9s\n", inputLine);
// More stuff here
} while( //some condition);
However if you edit your code and remove * you get answer, but normal array deprecated, nowdays, programmers use vector, normal array in C not safe :
#include <iostream>
#include <vector>
using namespace std;
int main() {
vector<string> inputLine;
You can define with every data type:
vector<int> myvar;
Or you can define multidimensional vector:
vector< vector <int> > myvar;
Related
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <conio.h>
char *preorden="GEAIBMCLDFKJH";//line 5
error in the above line
char *inorden="IABEGLDCFMKHJ";//line 6
error in this line
char *postorden;
error in this line
void post(char *pre, char *in, char *pos,int n)
{
int longIzqda;
if(n!=0){
pos[n-1]=pre[0];
longIzqda=strchr(in,pre[0])-in;
post (pre+1,in,pos,longIzqda);
post (pre+1+longIzqda,in+1+longIzqda,pos+longIzqda,n-1-longIzqda);
}
}
int main(int argc,char *argv[])
{
int aux;
aux=strlen(preorden);//convert to string
postorden=(char *)malloc(aux*sizeof(char));//use of malloc function
if (postorden){
printf("The preorden is: %s\n",preorden);
printf("The inorden is: %s\n",inorden);
post(preorden,inorden,postorden,aux);
postorden[aux]='\0';
printf("The postorden calculated is: %s\n",postorden);
free(postorden);
}
else{
fprintf(stderr,"Whithout memory\n");
return 1; // return 1
}
return 0;
}
the error is in the line 5 and 6
the compiler says:
deprecated conversion from string constant to 'char*' [-Wwrite-strings]
There are few issues with your code, firstly this
char *preorden="GEAIBMCLDFKJH";//line 5
forces compiler to warn you like below if compiled with -Wwrite-strings flags in C
deprecated conversion from string constant to 'char*'
[-Wwrite-strings]
because the string literal GEAIBMCLDFKJH stored in read only section of primary memory i.e pointer where it points, that contents is read only, hence instead of char* use const char*. for e.g
char *preorden = "GEAIBMCLDFKJH";/* preorden is normal pointer but "GEAIBMCLDFKJH" is read only, hence error */
And
const char *preorden = "GEAIBMCLDFKJH"; /* const char *ptr means ptr contents is read only */
Secondly, here
postorden=(char *)malloc(aux*sizeof(char));//use of malloc function
casting of malloc result is not required as malloc() return type is void* which is automatically and safely promoted to any other pointer type, Read Do I cast the result of malloc?. for e.g
postorden = malloc(aux * sizeof(*postorden));//use of malloc function
Also here(this point is about wrong comment on below line, please don't mind)
aux=strlen(preorden);//convert to string
strlen(preorden) returns the length of string pointed by preorden and gets assigned to aux not as written in comments(convert to string).
And change the post() definition to
void post(const char *pre, const char *in, char *pos,int n) {
/* some code*/
}
The message “deprecated conversion from string constant to 'char*' [-Wwrite-strings]” arises because the code was compiled as C++ code, which has different rules about string literals and pointer conversions from C.
This can be fixed by compiling the code as C code or worked around by inserting an explicit cast to char *.
I am working on a project that requires that I make an array of a certain structure type. The structure looks like this:
typedef struct
{
char array[30];
int num;
}structure
I have declared an array of these structures like this:
structure struct1[5];
I passed this structure array to a function which fills the first four elements with a string and a number. The fifth is left blank. Later in the program, I pass the array again and try to set the string in the structure to a user determined string using gets(). I am getting this error:
438:19: error: incompatible types when assigning to type 'char[30]' from type 'char *'
If I need to provide more clarification, please tell me.
Thanks!
EDIT: Here is what I am doing:
typedef struct
{
char array[30];
int num;
}structure;
void fillStructs(structure[]);
void userEditStructs(structure[]);
main()
{
structure myStructs[5];
fillStructs(myStructs);
userEditStructs(myStructs);
return 0;
}
void fillStructs(structure s[])
{
//code to fill myStructs elements 0-3.
}
void userEditStructs(structure s[])
{
char newArr[30];
int newNum;
printf("Please enter your string:\n\nEntry:\t");
gets(newArr);
printf("Please enter your number:\n\nEntry:\t");
scanf("%i", newNum);
s[5].array = newArr;
s[5].num = newNum;
}
you are doing something like this
char a[20];
a = "bla";
you cant do this.
do strcpy(a,"bla"); instead. ( #include <string.h> )
Without looking at the code you are probably trying to do something like:
struct[4].array = gets(<unknown>);
which won't work, as you can't assign the returned char* from gets to an array as the compiler says. You are also using gets, which is strongly discouraged as it performs no bounds checking. Instead, do the following:
fgets(struct[4].array, sizeof(struct[4].array), stdin);
which will do proper bounds checking.
I'm new to the world of C programming, and I as trying to code a primitive, terminal-based version of the "Hangman" game.
One of the steps doing this (or at least the way I am working on), is to create a second char array (next to the original char array that stores the word one needs to guess), filled with "*" for every Char of the original array, and display it. Although the rest of the programming part is not there yet (since I am not finished with it yet), I doubt it is relevant for now (however I allready know how to proceed, that is if I weren't bothered by some error-messages)....
Here's the code I have so far:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
void hiddenArray(char array[]);
char *secretWord;
char *arrayCover;
char letter;
int main(int argc, char *argv[]){
secretWord = "Test";
printf("Welcome to the Hangman!\n\n");
hiddenArray(secretWord);
printf("What is the hidden Word?: %c\n", *arrayCover);
printf("Guess a letter\n");
scanf("%c", &letter);
}
void hiddenArray(char array[]){
int size,i;
size = sizeof(*array);
for (i=0;i<size-1;i++){
*(arrayCover+i) = "*";
}
}
Now I have two issues... the first one:
I don't understand the error message I am getting after compilation:
pendu.c:41:19: warning: incompatible pointer to integer conversion assigning to 'char' from 'char [2]' [-Wint-conversion]
*(arrayCover+i) = "*";
^ ~~~
1 warning generated.
And my second question: the second Array created, filled with "*" is not being displayed, what did I do wrong?
I'd love for some help, cheers!
Your program have some errors to be corrected .
1) Your *arraycover is pointing to some unknown value, You have not initialized it.
2) sizeof(*array) should be sizeof(array)
3) *(arrayCover+i) = "*" should be *(arrayCover+i) = '*';
I suggest you not to create too many global variables when you dont really need them
Create char secretWord[100] = "Test" instead of char *secretWord
Try this
size = sizeof(arrayCover)/sizeof(char);
for (i=0;i<size-1;i++){
*(array+i) = '*';
}
Despite of all these, I think you need to allocate memory for arrayCover
arrayCover = malloc(sizeof(char) * number_of_elements);
"*" is a string. Use '*' instead to get the char.
"*" is a string which also contains \0 also, resulting to 2 characters. Instead use '*' which is just a character.
The function hiddenArray has the formal argument array which need to be used locally instead of arrayCover as below,
void hiddenArray(char array[]){
int size,i;
size = sizeof(array)/sizeof(array[0]);
for (i=0;i<size-1;i++){
*(array+i) = '*'; /* Or array[i] = '*' */
}
}
This question already has answers here:
Passing two-dimensional array via pointer
(9 answers)
Closed 9 years ago.
I am trying to make function that fills array of strings with lines from file, but the compiler (GCC) still giving me a warning. Than if I try to run compiled app, it gives me "Segmentation fault" error
Source code:
main
#include <stdio.h>
#include "getAdresses.h"
int main(int argc, char **argv){
char adresses[1024][128];
getAdresses(adresses);
printf("%s", adresses[1]);
}
getAdresses
include <stdio.h>
int getAdresses(char **adresses){
FILE *fr;
fr = fopen("adresses", "r");
int i = 0;
while(adresses[i-1][0] != EOF){
fscanf(fr, "%s\n", &adresses[i]);
i++;
}
}
It's giving me this error:
main.c: In function ‘main’:
main.c:9:2: warning: passing argument 1 of ‘getAdresses’ from incompatible pointer type [enabled by default]
In file included from main.c:3:0:
getAdresses.h:1:5: note: expected ‘char **’ but argument is of type ‘char (*)[128]’
First of all you have done the typical mistake: char ** is not the same as a char (*)[128]. The latter is the type of adresses.
You get the segmentation fault when you try to dereference it in line
while(adresses[i-1][0] != EOF)
Aside from the fact that adressing [i-1] for i = 0 will give you bad results you should define your function as
int getAdresses(char (*adresses)[128])
to be able to pass your two dimensional array correctly and fscanf should scan into your actual line buffer and if you are reading line by line use fgets:
while(fgets(adresses[i], 128, fr)) i++;
when you pass adresses[1024][128] to a function, the compiler will only do one step of decay, and the parameter type should be char (*)[128] instead of char ** .
You also need to pass its first dimension to the function, which means
int getAdresses(char (int (*arr)[128], int x)
check the c-faq for a detailed explanation for this issue.
The array you allocated declaring
char adresses[1024][128]
is actually a char* (it isn't but it's much closer to it than to char**).
The double diemension accessor is actually just a syntactic sugar for [x+y*width].
To pass the array as a parameter use:
int getAdresses(char (*adresses)[128])
I just started to look at C, coming from a java background. I'm having a difficult time wrapping my head around pointers. In theory I feel like I get it but as soon as I try to use them or follow a program that's using them I get lost pretty quickly. I was trying to follow a string concat exercise but it wasnt working so I stripped it down to some basic pointer practice. It complies with a warning conflicting types for strcat function and when I run it, crashes completly.
Thanks for any help
#include <stdio.h>
#include <stdlib.h>
/* strcat: concatenate t to end of s; s must be big enough */
void strcat(char *string, char *attach);
int main(){
char one[10]="test";
char two[10]="co";
char *s;
char *t;
s=one;
t=two;
strcat(s,t);
}
void strcat(char *s, char *t) {
printf("%s",*s);
}
Your printf() should look like this:
printf("%s",s);
The asterisk is unnecessary. The %s format argument means that the argument should be a char*, which is what s is. Prefixing s with * does an extra invalid indirection.
You get the warning about conflicting types because strchr is a standard library routine, which should have this signature:
char * strcat ( char * destination, const char * source );
Yours has a different return type. You should probably rename yours to mystrchr or something else to avoid the conflict with the standard library (you may get linker errors if you use the same name).
Change
printf("%s",*s);
to
printf("%s",s);
The reason for this is printf is expecting a replacement for %s to be a pointer. It will dereference it internally to get the value.
Since you declared s as a char pointer (char *s), the type of s in your function will be just that, a pointer to a char. So you can just pass that pointer directly into printf.
In C, when you dereference a pointer, you get the value pointed to by the pointer. In this case, you get the first character pointed to by s. The correct usage should be:
printf( "%s", s );
BTW, strcat is a standard function that returns a pointer to a character array. Why make your own?
Replacing *s with s won't append strings yet, here is fully working code :
Pay attention to function urstrcat
#include <stdio.h>
#include <stdlib.h>
/* urstrcat: concatenate t to end of s; s must be big enough */
void urstrcat(char *string, char *attach);
int main(){
char one[10]="test";
char two[10]="co";
char *s;
char *t;
s=one;
t=two;
urstrcat(s,t);
return 0;
}
void urstrcat(char *s, char *t) {
printf("%s%s",s,t);
}
pointers are variable which points to address of a variable.
#include "stdio.h"
void main(){
int a,*b;
a=10;
b=&a;
printf("%d",b);
}
in the follwing code you will see a int 'a' and a pointer 'b'.
here b is taken as pointer of an integer and declared by giving'' before it.'' declare that 'b' is an pointer.then you will see "b=&a".this means b is taking address of integer "a" which is keeping value 10 in that particular memory and printf is printing that value.