Discussion :: Command Line Arguments - Find Output of Program (Q.No.1)
| A.Rajesh said: (Sep 13, 2010) | |
| I think here directory #include<stdlib.h> shouldn't mentioned. But I tried in GCC compiler(linux) i got the out put like this rajesh@rajesh-laptop:~$ vim prog.c rajesh@rajesh-laptop:~$ cc prog.c rajesh@rajesh-laptop:~$ ./a.out one two three o rajesh@rajesh-laptop:~$ |
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| Seenu said: (Sep 17, 2010) | |
| argv[0] = myprog argv[1] = one So **++argv=> argv[1]. And %c is output so first char of one ==> o. |
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| Anand Shankar Jha said: (Jan 6, 2011) | |
| I think seenu's logic is good enough to answer the question. Good seenu. since ++ is prefix with argv hence pre increament happens in argument and the the array is char type, so, it stores one char at a time. So, Answer must be C i.e. "o". |
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| Navneet Agarwal said: (Feb 9, 2011) | |
| Seenu answer is good and understood. | |
| Murthy said: (Mar 18, 2011) | |
| Please give me brief description. | |
| Khagesh Gupta said: (Mar 28, 2011) | |
| Real thing is here *argv=pointer to "myprog"(=base address) and equivalent to argv[0] then preincrements *++argv=pointer to "one"(more specifically pointer to "o") equivalent to argv[1] **++argv=value at this address thus ans is o |
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| Satya said: (Jun 7, 2011) | |
| Senu answer is good. | |
| Gowri Shankar said: (Aug 2, 2011) | |
| Seenu your answer is acceptable, but Khagesh Gupta is correct. | |
| Ashok said: (Oct 7, 2011) | |
| Seenu answer is correct. | |
| Shambhu said: (Nov 10, 2011) | |
| I think seenu is correct because argv pointing to the first element that is myprog, when it is incremented (++ has higher priority than *), now it pointed to "one",but we have agian one * that will point to 'o'. | |
| Vivek said: (Dec 4, 2011) | |
| Seenu your answer is corretct n understand. Good explain. | |
| Rupinderjit said: (Dec 6, 2011) | |
| Here one thing to be taken into consideration is PRECEDENCE.Since ++ has high precedence over *(indrection),so pre-incrementation takes place first.and *argv==**argv,since both pointing to same address. Khagesh Gupta alucidation in not so correct. Moreover if we use array indexing here,then ++argv[i] means increment the value at argv[i] and display it.Andargv[i]++ means,value at argv[i] then increment the address using post increment protocol. |
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| Shahul Hameed P (Sinuxcreation@Gmail.Com) said: (Jan 5, 2012) | |
| **++argv will point to argv[1] If again we do that (ie, **++argv) it will point to argv[2] Here %c is the control string,So output will be the values present at their initial location[ie,if pointing to argv[1] it will show the character present at 0th location of argv[1]). |
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| Deepak Kumar said: (Mar 4, 2012) | |
| argv--->is the address of 1st element of argv[] *argv---->points to the first element of argv[] *++argv--->points to the next element of argv[],and its also points to 1st element of *++argv ie j of jan **++argv---->print j |
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| Cherry said: (Jul 1, 2012) | |
| Seenu's answer is absolutely right. i.e., Here argv[0]=myprog argv[1]=one argv[2]=two argv[3]=three **++argv[] it is pre-increment, initially it shows "myprog" Now, After the pre-increment it shows "one" But given printf("%c",**++argv[]); c for character so that at a time it stores only one character i.e "o" If "%s" is there in printf then it stores whole string"one" I hope you understand. |
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| Sadanand said: (Jul 11, 2012) | |
| **++argv means argv[0] but this will give first chareacter of argv[0].. In this case the value is 0. If you want second character of that same string you can write like this *(*(argv+0)+2) |
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| Vineet said: (Mar 18, 2013) | |
| argv have address of argv[0]; argv[0]="myprog"; argv[1]="one" . . . and so on... *(*(++argv))); ++argv means address of argv[1] and now value at argv[1] is address of o of(one). then again value at that address is o. |
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| Shalini said: (Nov 5, 2013) | |
| **argv is pointer to array of pointers. *argv---->argv[0]---->myprog. *argv+1-->argv[1]---->one. . . . argv having base address of "myprog". ++argv having base address of "one". *++argv(dereference) prints "one". **++argv(two times dereference) prints 'o'. |
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| Sunaina said: (Jan 29, 2014) | |
| Another question is : What will be the output of the program (myprog.c) given below if it is executed from the command line? cmd> myprog friday tuesday sunday /* myprog.c */ #include<stdio.h> int main(int argc, char *argv[]) { printf("%c", *++argv[1]); return 0; } For this answer is "r". Can anyone explain difference in these ? As the confusion is that in this question after ++ we move to next string and in another question as I mentioned above we move to one character.... why? |
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| Akhilesh said: (Apr 20, 2014) | |
| *argv=argv[1]=argv+1. So, ++argv[1]=*argv. i.e, *(++(*argv)) = here value is incremented so it will Print next character provided %c should be there in printf. |
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| Swapnil said: (Jun 13, 2016) | |
| Thanks @Cherry. | |
| Priyanka said: (Sep 2, 2016) | |
| Consider a pointer *p for storing single dimension array. *++p points to next element in the array. In the case of 2-dimensional array **p is used and **++p points to next word. Finally coming to answer **argv[] is a 2-dimensional representation of command line arguments and, **++argv[] points to next word. %c is used for printing character at that position. So, the answer is "o". |
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| Goriparthi. Udaya Lakshmi said: (Jan 7, 2018) | |
| How the answer comes o? | |
| Nikhil N said: (Jan 25, 2018) | |
| Perfect answer @Seenu. | |
| Sanjana said: (Jul 16, 2018) | |
| But argv[0] must be a program name right? | |
| Sravya said: (Aug 27, 2018) | |
| Good explanation @Cherry. | |
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