execv system call not running as desired - linux

Question

I use Linux and while compiling any c or cpp file, I use gcc or g++ respectively in terminal.
Common syntax : g++ program.cpp
But now I wish to compile files using flags.
Eg: g++ -Wall -Wextra -std=c++11 program.cpp
I will use more 10 flags to compile my program. But I don't want to remember and type that while compiling in terminal.
Now I wish to create a c program involving syscalls (exec) to get my job done using below syntax:
./compile program.cpp
But there's some problem while using exec in my below code
#include<fcntl.h>
#include<unistd.h>
#include<stdlib.h>
int main(int args, char* argv[]){
char* arguments[10]={"-std=c++11","-Wall","-Wextra","-pedantic","-Wshadow","-fsanitize=address","-fsanitize=undefined","-fstack-protector"}; //consists of flags with which i will compile the program passed as argument
printf("%s\t %s",argv[0],argv[1]);
if(args==2){
arguments[8]=argv[1];
arguments[9]=(char*)NULL;
}else{
printf("only one argument allowed!");// just to check if i pass arguments correctly
}
printf("%s\t %s",arguments[8],arguments[9]);// just to check if my arguments array is correct
if(execv("/bin/g++",arguments)==-1){ // to my suprise this line runs before above printing lines. What is the reason/solution?
perror("execv failed!");
exit(1);
}
return 0;
}
The above code compiles successfully without error.
But I think execv runs even before I insert passed argument in argument array.
Because of which, program runs with error execv failed: no such file or directory
Followed by the printfs.
Please tell me where I went wrong.

Answer 1

So I finally solved the ambiguity in the above code. I made two radical changes to my code.
Instead of directly assigning argument strings while declaration of string array,
char* arguments[10]={"-std=c++11","-Wall","-Wextra","-pedantic","-Wshadow","-fsanitize=address","-fsanitize=undefined","-fstack-protector"};
I chose to just assign strings one by one.
char* arguments[10];
arguments[0]="g++";
arguments[1]="-Wall";
arguments[2]="-Wextra";
and so on
And this fixed the segmentation faults in my code.
This time I used execvp() instead of execv() system call because of which I don't need to explicitly declare full path to the command usr/bin/g++ and so on. In execvp only command name is enough.
So my new code looks like this:
#include<stdio.h>
//#include<fcntl.h>
#include<unistd.h>
#include<stdlib.h>
int main(int args, char* argv[]){
char* arguments[10];
//printf("%s\t %s\n",argv[0],argv[1]);
arguments[0]="g++";
arguments[1]="-Wall";
arguments[2]="-Wextra";
arguments[3]="-pedantic";
arguments[4]="-Wshadow";
arguments[5]="-fsanitize=address";
arguments[6]="-fsanitize=undefined";
arguments[7]="-fstack-protector";// to add more flags make changes in this array.
if(args==2){
arguments[8]=argv[1];
arguments[9]=(char*)NULL;
if(execvp(arguments[0],arguments)==-1){
perror("execv failed!");
exit(1);
}
}else{
printf("->Only one argument(c/cpp file) allowed.\n->Requtired syntax: ./compile program.cpp\n");
}
return 0;
}
Another question I had was that all printfs that were before execv() system call would get printed only after execv() got executed. And as #MYousefi Sir commented, it was because of buffer not being full. And as suggested, adding "\n" in printfs solved the problem.