rds95@leah.Albany.Edu (Robert Seals) (08/10/89)
Hello frenz, is it workable to pass only the base address of an array
to "?scanf" and have it convert into successive memory locations?
int d[4];
scanf("%d %d %d %d", d);
I guess the question is whether "scanf" uses the format string or
the number of arguments to determine how many thingies to convert.
So, what is it?
robgwyn@smoke.BRL.MIL (Doug Gwyn) (08/10/89)
In article <1949@leah.Albany.Edu> rds95@leah.Albany.Edu (Robert Seals) writes: >int d[4]; >scanf("%d %d %d %d", d); Since the array name "d" is converted to "pointer to element d[0]" before being passed to scanf(), the question answers itself. d[0] is a single int. scanf() can't tell the difference between ways of producing pointers-to-int that it is fed as arguments; they all look the same to it. >I guess the question is whether "scanf" uses the format string or >the number of arguments to determine how many thingies to convert. The format string and the input data determine how many items scanf() actually converts. Correct usage requires that you provide one pointer argument for each potentially converted item, i.e. the remaining arguments must exactly correspond to the format string.
jeffa@hpmwtd.HP.COM (Jeff Aguilera) (08/11/89)
> Hello frenz, is it workable to pass only the base address of an array > to "?scanf" and have it convert into successive memory locations? > > int d[4]; > > scanf("%d %d %d %d", d); > No. Use scanf("%d %d %d %d", d, d+1, d+2, d+3); For each valid conversion specification other than %%, at least one pointer is consumed from the stack. You pass one pointer, but ask for four conversions. Expect a hung system or dumped core, depending upon available memory protection.
foessmei@lan.informatik.tu-muenchen.dbp.de (Reinhard Foessmeier) (08/11/89)
In article <1949@leah.Albany.Edu> rds95@leah.Albany.Edu (Robert Seals) writes: >...is it workable to pass only the base address of an array >to "?scanf" and have it convert into successive memory locations? > >int d[4]; > >scanf("%d %d %d %d", d); > >I guess the question is whether "scanf" uses the format string or >the number of arguments to determine how many thingies to convert. "scanf" uzas la formatan vicon; "scanf" uses the format string; it ^gi ne scias la nombron de argumentoj; doesn't know about the # of args; but tamen via propono ne funkcias. what you propose doesn't work all the La kawzo estas, ke "scanf" volas same. The reason is that "scanf" wants propran adreson por ^ciu legata an address of its own for every datum dateno. ^Car la adreso de unu read. Since the address of a single elemento de estas distingebla de element is not distinguishable from the la adreso de vektoro, "scanf" ne address of an array, "scanf" would be scius kion fari el at a loss with something like scanf(" %d %d %d", d1, d2); d1 kaj d2 povus esti deklaritaj kiel d1 and d2 might be declared as int d1[2], d2[1]; aw or int d1[1], d2[2]; "scanf" absolute ne scias, kiu el "scanf" has no way of telling which of la du deklaracio antawiris. these declarations was used. Reinhard F\"ossmeier, Technische Univ. M\"unchen | Vivu foessmeier@infovax.informatik.tu-muenchen.dbp.de | la gefiloj [ { relay.cs.net | unido.uucp } ] | de niaj gepatroj!
chris@mimsy.UUCP (Chris Torek) (08/11/89)
In article <1949@leah.Albany.Edu> rds95@leah.Albany.Edu (Robert Seals) writes: >Hello frenz, is it workable to pass only the base address of an array >to "?scanf" and have it convert into successive memory locations? > >int d[4]; > >scanf("%d %d %d %d", d); Easy, yes; dumb, no: but the answer is no. >I guess the question is whether "scanf" uses the format string or >the number of arguments to determine how many thingies to convert. >So, what is it? It uses the format. In particular, the proposed ANSI C standard says that printf et al. can be given `extra' arguments, so these clearly must use the format rather than (or perhaps in addition to) the number and types of arguments; and given that, it is likely that the same will be true of scanf, whether or not it is required (I cannot recall offhand). In any case, even if it used the number of arguments, scanf("%d %d %d %d", d); would still pass only two arguments, one being a pointer to char that points to the first `%', and the other being a pointer to int that points to d[0]. -- In-Real-Life: Chris Torek, Univ of MD Comp Sci Dept (+1 301 454 7163) Domain: chris@mimsy.umd.edu Path: uunet!mimsy!chris
dkelly@npiatl.UUCP (Dwight Kelly) (08/11/89)
In article <1949@leah.Albany.Edu> rds95@leah.Albany.Edu (Robert Seals) writes: >Hello frenz, is it workable to pass only the base address of an array >to "?scanf" and have it convert into successive memory locations? > >int d[4]; > >scanf("%d %d %d %d", d); > Not possible. Scanf reads the variable addresses off of the stack. In your example, it would get the address of d[0] and then get three random addresses for the next three %d. One solution is: scanf("%d %d %d %d", &d[0], &d[1], &d[2], &d[3]); or: int i; for (i=0; i<4; i++) scanf("%d", &d[i]); -- Dwight Kelly UUCP: gatech!npiatl!dkelly Director R&D AT&T: (404) 962-7220 Network Publications, Inc 2 Pamplin Drive Lawrenceville, GA 30245 Publisher of "The Real Estate Book" nationwide!