Discussion :: Command Line Arguments - Find Output of Program (Q.No.4)
| Aishwarya said: (Dec 3, 2010) | |
| The 'argc' contains count value and argv[1] contains "Good". On printing argc,argv,its o/p will be 3 Good. Because count value is 3 here. |
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| Xyz said: (Dec 27, 2010) | |
| But sample is file name right arguments are given after file name is specified then good =0,morning =1 .. I don't understand if sample is argument here. | |
| Sundar said: (Mar 22, 2011) | |
| The given answer is correct. I have tested it. argc - will contain 3 (count of arguments argv[]) argv[0] = C:\TURBOC\SAMPLE.EXE argv[1] = Good argv[2] = Morning Therefore, answer is "3 Good" |
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| Prasastha said: (Jun 29, 2011) | |
| argc=argument count= sample,good,morning=3 argv[0]=by default it stores base address= it stores file name=sample argv[1]=good argv[2]=morning %d argc=3 %s argv[1]=good |
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| Vidhya Balan said: (Jan 6, 2012) | |
| Tell me - what is command line argument? | |
| Sundar said: (Jan 6, 2012) | |
| @Vidhya Balan Assume your current working directory in DOS is "C:\TURBOC" and your C program name is SAMPLE.C (which is in c:\turboc directory). If you compile the program successfully, an EXE file SAMPLE.EXE will be created in your current directory i.e "c:\turboc". You can run it by simply typing its name as given below. C:\TURBOC>sample.exe Good Morning <press-enter-key> (or) C:\TURBOC>SAMPLE Good Morning <press-enter-key> Here "sample" is the 'exe-file-name-or-command' and "Good Morning" is the arguments supplied to the exe-file-or-command called "sample" which is created by compiling the C program "sample.c". Hope this will help you. Have a nice day! |
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| Jayant Jadhav said: (Feb 20, 2012) | |
| argc which count number of word or characters separated by space and it holds int value. In this example three words are their including file name.Hence argc print 3. argv[] this is argument vector array which holes all arguments in string format and store from starting index with 0. argv[0]=sample argv[1]=Good..... |
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| Vijay said: (Mar 30, 2012) | |
| Can any one explain how actually memory allocation is done in command line argument. | |
| Jogamohan Medak said: (Oct 20, 2012) | |
| argv[0]=Base address=Sample argv[1]=Good argv[2]=Morning %d argc=3 because (argv[0],argv[1],argv[2] only three arguments argv[]) %s argv[1]=Good |
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| Abir Sen said: (Oct 25, 2013) | |
| In linux, argc = 3 because no of arguments that are passed is 3 in the program. arv[0] = ./a.out, argv[1] = Good, argv[2] = Morning. |
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| Padmapriya said: (Sep 14, 2014) | |
| But I have a doubt guys. Usually the count value will start with 0, but how here started with 1. Is that the rule for argc[]. Please explain. | |
| Naga Vikas said: (Mar 24, 2015) | |
| Does this depends on the compilers? Because I get it as 4 Sample. 0--Path. 1--Sample. 2--Good. 3--Morning. Anyone please tell me? |
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| Zara said: (Feb 15, 2016) | |
| Sample Good Morning are three words, so argc will total words like sample = 1 good = 2 and morning = 3 that is total 3 counts and argv contains argv[0] = sample, argv[1] = good, and argv[2] = morning. So output will be at printf ("%d %s", argc, argv[1]); 3 count at argc, and good at argv[1] that is 3 Good. |
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| Prash... said: (Dec 8, 2016) | |
| Thanks @Sundar. | |
| Garima said: (May 8, 2017) | |
| Why it count? Please explain me. | |
| Datts said: (Nov 11, 2018) | |
| After a first value insertion count is automatically incremented i.e. count++. | |
| Meera said: (Aug 2, 2020) | |
| I am not understanding this, Please anyone help me. | |
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