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can any one help me in my code
#include <stdio.h> #include <stdlib.h> #include <string.h> int longage2(char phe[ ]){ if (strlen(phe)==1||strlen(phe)%2!=0||phe[strlen(phe)]=='a'){ return 0; } for(int i=0;i<strlen(phe);i++){ if(i%2==0){ if(phe[i]!='a'){ return 0; }}else{ if(phe[i]!='b'){ return 0;}} }return 1;} void longage4(char phe[]){ int a;char ph1[10]; for( int i=0;i<=strlen(phe)-1;i++){ ph1[i]=phe[i]; printf("%d\n",longage2(ph1)); if(longage2(ph1)==1){ a=i; }} printf("a=%d\n",a); char ph2[a]; for(j=0;j<=a;j++){ ph2[j]=ph1[j]; } printf("%s\n",ph2); printf("%d",longage2(ph2)); } int main() {char phe[20]; printf("donnez un mot\n"); scanf("%d",phe); longage4(phe); return 0; }
11 Answers
+ 5
mar ia the problem is that ph2 has no terminating character '\0'
Also if you want ph2 to contain ph2[3] you should declare char ph2[4]; so char ph2[a+1];
adding another space for \0 it become:
char ph2[ a + 2 ]
for(j=0; j <= a; j++) {
ph2[i] = ph1[i];
}
ph2[a+1] = '\0';
+ 4
The scanf in the main() should have %s
And you need to declare int j; into longage4()
+ 4
Can you tell me what to enter as input?
+ 3
Thank you very much for helping me. I don't know😦 how to express my thanks to you😀😊
+ 1
I fex it but wan yo but (donne un mot and you raet abab l hav proplem in resolt of line 32
+ 1
He raet to me "abab?"
I didnt no whay but (?)qustshen marc
+ 1
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int longage2(char phe[ ]){
if (strlen(phe)==1||strlen(phe)%2!=0||phe[strlen(phe)]=='a'){
return 0;
}
for(int i=0;i<strlen(phe);i++){
if(i%2==0){
if(phe[i]!='a'){
return 0;
}}else{
if(phe[i]!='b'){
return 0;}}
}return 1;}
void longage4(char phe[]){
int a;char ph1[10];int j;
for( int i=0;i<=strlen(phe)-1;i++){
ph1[i]=phe[i];
if(longage2(ph1)==1){
a=i;
}}
char ph2[a];
for(j=0;j<=a;j++){
ph2[j]=ph1[j];
}
printf("%s\n",ph2);
}
int main()
{char phe[20];
printf("donnez un mot\n");
scanf("%s",phe);
longage4(phe);
return 0;
}
+ 1
thes is beter an clen and the input is abab and mast be the reasolte is abab
+ 1
The program works like this
When we enter (ab)^n n>0
the reasolt is (ab)^n
wen we but like exp : aba the resolt shod be :ab
He is looking for (ab)^n Consecutive and he Display it
0
Hug me 🤗🤗🤗
0
Hhhhh😊