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I'm trying to convert a char array to upper case in C. But it still prints out in lower case. The logic in conversion function looks okay to me. I don't know if the problem could be that the input I'm passing to this function is member of a structure variable.
This is the structure whose member I'm passing to conversion function:
typedef struct AESParams {
unsigned char output[192];
} AESParams;
My conversion fucntion is shown below:
void stringUpr(char *s)
{
int i=0;
while(s[i]!='\0')
{
if(s[i]>='a' && s[i]<='z'){
s[i]=s[i]-32;
}
++i;
}
}
I call conversion function as follows:
AESParams parameters;
stringUpr(parameters.output);
If I print the output after calling "stringUpr" function, the value is still in lower case. Can anyone tell me what might be causing the issue here?
TIA...
Update: Code that writes value to output.
EVP_EncryptUpdate(&ctx, parameters->output, &outLen, input, length);
// This call happens in some other file. parameters is passed to the function that calls this function and after computation parameters is passed back to the place where I'm calling stringUpr().
I'm confident that this lines works and it gives the correct results after computation. It's just that I want to take this value and convert to upper case and write to a file.
Your program assumes that it's running using an ASCII character set, which is not guaranteed. Use the standard functions defined in ctype.h
void stringUpr(char* s)
{
int i = 0;
while(s[i] != '\0')
{
s[i++] = toupper((unsigned char)s[i]);
}
}
Related
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The code which does not work. In function main after initializing variables by the user, it goes to if part and after that crash happens. This program's goal is to impact the repetitive characters in the string. So I decided to have a pointer in a function f1 and the pointer should point to 0th char of the array kar[]
then compare the 0th char to 1st char but the program crashes exactly after the last scanf("%s",kar[1000]).
where is the problem?
#include <stdio.h>
char f1(char* p)
{
int i=0,c=1;
while(*(p+i))
{
if(*p==*(++p))
{
c+=1;
p++;
i++;
continue;
}
else
{
if(c>1)
{
printf("%c%d",*p,c);
}
else
{
printf("%c",*p);
}
}
i++;
}
}
int main()
{
int n,i,m;
char kar[1000];
scanf("%d",&n);
for(i=1;i<=(2*n);++i)
{
scanf("%d",&m);
scanf("%s",kar[1000]);
if(m==1)
{
f1(kar);
}
}
}
This
scanf("%s",kar[1000]);
should trigger a compiler warning, as you pass a char where a char* is expected. You pass the 1001st element of kar (which is out of kar's bounds, BTW). In C array indices start with 0 for the 1st element.
To scan into kar, just pass the address of kar's 1st element.
scanf("%s", &kar[0]);
As arrays get decayed to the address of their 1st element when being passed to a function the following statement is equivalent
scanf("%s", kar);
Please note that the above two statements very well allow the user to enter more characters then kar can hold and with this make scanf() overflow kar, that is write beyond it bounds, invoking undefined behaviour, which might crash your program or whatever ...
To avoid this tell scanf() the maximum to scan by doing:
scanf("%999s", kar);
The maximum for chars a strings can hold in C is always one less then defined as C strings need one more char to have their end marked by a '\0'terminator.
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This function make a strange error after using it several times and I really can't understand the reason behind it.
char *get_range(char *str,int min,int max){
char *_res=(char *)malloc(sizeof(str));
int cur=0;
while (min<max){
_res[cur]=str[min];
min++;
cur++;
}
return _res;
}
The problem is that after using this function several times, the output comes with additional chars and I don't understand why.
Notice: The additional chars are allway used returned by the function beffor
char *_res=(char *)malloc(sizeof(str));
is wrong. sizeof(str) is measuring the size of a char pointer. This is either 4 or 8 (typically) depending on your system (32 or 64 bit).
You need
char *_res=(char *)malloc(strlen(str) + 1);
strlen returns the number of characters in the string, and you need to add 1 for the terminating 0;
Second you have to add a terminating zero at the end, do:
_res[cur] = '\0';
before returning
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I have the following scenario where any input string will be converted into an integer.
Example:
result = get_integer_from_string("100");
result == 100; // true
How can I write this function without using any libraries? I am able to do it by using libraries.
Follow these Steps:
Parse the input string.
Check if the character is a digit or not.
Use some logic to convert digit in character format to integer format.
Also, you can implement Exception in case, input is not integer string.
I cannot tell you the code, it would not help you in learning, try implementing code yourself, it is very easy!!
Loop over the string, from the end to the start. Get each digit, and convert it to a decimal value. Multiply the first (in the backward loop) by 1 and store the result. Multiply the second by 10 and add to the result of the previous. And so on.
This is very prone to error conditions, but should work if string is valid integer:
int str2int(const char* str) {
int result = 0;
char* p = str;
for (;;) {
char c = *p++;
if (c < '0' || c > '9')
break;
result *= 10;
result += c - '0';
}
return result;
}
It has behavior close to atoi() - stop processing on any non-digit, and return 0 for empty input.
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I am trying to program an encrptor and decryptor. My problem is a pointer that I assigned value first still has the same value as the second. I tried to use free but the problem is still the same.
For example, I typed aslkdjlasc(sample only)
and the output would be:
helloworld
I try to run the program again, then I type daskjda
the output would be like
doctorxRLD
RLD is from the past value of this pointer. It's supposed to be 7 characters only, but since helloworld is 10, the first 7 letters are replaced by the encryption but the last 3 characters are still printed.
What do I do?
UPDATE: HERE IS PART OF THE CODE:
void encrypt(char *crypt)
{
char *plaintext,*encryption,slash=0x2F;
int i,j,k,flags,f;
encryption=(char *)malloc(sizeof(int));
plaintext=(char *)malloc(sizeof(int));
printf("Enter plaintext, spaces,commas,and dots should be represented as /:\n");
scanf("%s",&*plaintext);
for(i=0;i<strlen(plaintext);i++)
{
j=1;flags=0;
while(j<53 && flags==0)
{
if(plaintext[i]==slash)
{
encryption[i]=slash;
flags=1;
}
if(plaintext[i]==crypt[j])
{
encryption[i]=crypt[j-1];
flags=1;
}
k=j+2;
j=k;
}
}
printf("%s",encryption);
free(encryption);
free(plaintext);
getch();
}
HERE IS THE MAIN
main()
{
char c;
int timer;
char crypt[53]="***i have hidden my encryption code***";
clrscr();
printf("Press e to encrypt, d to decrypt, ESC to exit.\n");
c=getch();
switch(c)
{
case(0x1b):
exit(0);
break;
case(0x64):
decrypt(crypt);
break;
case(0x65):
encrypt(crypt);
break;
default:
printf("INVALID. FORCE EXIT IN 3 SEC0NDS.");
delay(3000);
exit(0);
}
getch();
}
In your code you are allocating integer size (4 bytes ) of memory for a string
When you do
plaintext=(char *)malloc(sizeof(int));
Then by doing this
scanf("%s",&*plaintext);
your are possibly scanning a string of size more than that four characters ( however you allocated only four bytes) also
scanf("%s",&*plaintext); is equivalent to scanf("%s",plaintext); ( with the previous statement you are adding unnecessary computations.
it's me. I got it already. Thanks to all your comments though some are harsh. haha
I refrained from using malloc because apparently DCoder points out I do not know how to use them.
Thanks Sanyam Goel I have fixed my scanf too.
I used only 2 pointers instead of 4. What I did was I instantiated them in the main function instead of in each of the decrpyt and encrypt functions. Like this:
main()
{
char c,*from, *to; ..........
void encrypt(char *crypt,char *plaintext,char *encryption)
void encrypt(char *crypt,char *ciphertext,char *decryption)
So when I call either of them, I just put:
case(0x64):
decrypt(crypt,from,to);
break;
case(0x65):
encrypt(crypt,from,to);
break;
And at the end of the switch in the main function:
free(from); from=NULL;
free(to); to=NULL;
So now I have eliminated unnecessary pointers and extra processes.
THANK YOU EVERYONE. :)
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Let us say that the English alphabets A to Z has value start from 1. A = 1, B = 2, C = 3 and so on. Write a program which calls a function which accepts the name of a person which is constant character array and returns an integer value with sum of the Alphapets. What is the benefit of passing the name of the person as const char array?
Suppose somebody else provides me a function that takes non-const char * and does the job. What the function is actually implemented is like this:
int get_int_sum(char *name)
{
int sum;
//codes to calculate sum of alphas
name[0] += 1;
//continue
return sum;
}
When I call the function using
char my_name[] = "Yu Hao";
int reuslt = get_int_sum(my_name);
Even if I got the result I want, my_name is changed to "Zu Hao" without my notice. However, if a function has a prototype of
int get_int_sum(const char*name)
I am sure that the string I passed will not be modified.
One advantage is that elements in a array are protected from changing its value.
For example, here is simple code.
int Your_function(const char * a)
{
a[3] = 'A'; // this statement causes compile error.
// do something
return 0;
}