HElp me, what's wrong with this code?

Question

#include <stdio.h>
#include <string.h>
int ltou(char str[100]);

int main() {
char str[100];
scanf("%s", str);

// Lower to upper case conversion
ltou(str[100]);
return 0;
}

int ltou(char str[100]){
int len= strlen(str);
for(int i=0; i<len; i++){
if(str[i]>='a' && str[i]<='z') str[i]= str[i]-32;
}
printf("\n%s", str);
return 0;
}
I'm getting the following error message

timeout: the monitored command dumped core
./Playground/runner.sh: line 84: 11 Segmentation fault timeout "$runtime" "$output" - < "/usercode/inputFile"
ion]
10 | ltou(str[100]);
| ~~~^~~~~
| |
| char
./Playground/file0.c:3:15: note: expected 'char *' but argument is of type 'char'
3 | int ltou(char str[100]);
| ~~~~~^~~~~~~~

Answer 1

Alexander Thiem Yes I checked any way can be used

Answer 2

HElp me, what's wrong with this code?

#include <stdio.h>
#include <string.h>
int ltou(char str[100]);

int main() {
char str[100];
scanf("%s", str);

// Lower to upper case conversion
ltou(str);
return 0;
}

int ltou(char str[100]){
int len= strlen(str);
for(int i=0; i<len; i++){
if(str[i]>='a' && str[i]<='z') str[i]= str[i]-32;
}
printf("\n%s", str);
return 0;
}

//this should work

Answer 3

when you call the function you should just give str:
ltou(str);
otherwise you just give the function element 100 of your string...

Answer 4

Puthur Harthik Yes it is just the same, but this was not the mistake...

Answer 5

thankyou @Alexander thiem and @puthiur harthik